a演算法c語言實現的

A. 用c語言或c++實現，用深度優先演算法，寬度優先演算法和A演算法求解把數碼問題，請高手幫忙寫出代碼

看著懂，寫出來就困難了

B. A*搜尋演算法的代碼實現(C語言實現)

用C語言實現A*最短路徑搜索演算法，作者 Tittup frog(跳跳蛙)。 #include<stdio.h>#include<math.h>#defineMaxLength100 //用於優先隊列（Open表）的數組#defineHeight15 //地圖高度#defineWidth20 //地圖寬度#defineReachable0 //可以到達的結點#defineBar1 //障礙物#definePass2 //需要走的步數#defineSource3 //起點#defineDestination4 //終點#defineSequential0 //順序遍歷#defineNoSolution2 //無解決方案#defineInfinity0xfffffff#defineEast(1<<0)#defineSouth_East(1<<1)#defineSouth(1<<2)#defineSouth_West(1<<3)#defineWest(1<<4)#defineNorth_West(1<<5)#defineNorth(1<<6)#defineNorth_East(1<<7)typedefstruct{ signedcharx,y;}Point;constPointdir[8]={ {0,1},//East {1,1},//South_East {1,0},//South {1,-1},//South_West {0,-1},//West {-1,-1},//North_West {-1,0},//North {-1,1}//North_East};unsignedcharwithin(intx,inty){ return(x>=0&&y>=0 &&x<Height&&y<Width);}typedefstruct{ intx,y; unsignedcharreachable,sur,value;}MapNode;typedefstructClose{ MapNode*cur; charvis; structClose*from; floatF,G; intH;}Close;typedefstruct//優先隊列（Open表）{ intlength; //當前隊列的長度 Close*Array[MaxLength]; //評價結點的指針}Open;staticMapNodegraph[Height][Width];staticintsrcX,srcY,dstX,dstY; //起始點、終點staticCloseclose[Height][Width];//優先隊列基本操作voidinitOpen(Open*q) //優先隊列初始化{ q->length=0; //隊內元素數初始為0}voidpush(Open*q,Closecls[Height][Width],intx,inty,floatg){ //向優先隊列（Open表）中添加元素 Close*t; inti,mintag; cls[x][y].G=g; //所添加節點的坐標 cls[x][y].F=cls[x][y].G+cls[x][y].H; q->Array[q->length++]=&(cls[x][y]); mintag=q->length-1; for(i=0;i<q->length-1;i++) { if(q->Array[i]->F<q->Array[mintag]->F) { mintag=i; } } t=q->Array[q->length-1]; q->Array[q->length-1]=q->Array[mintag]; q->Array[mintag]=t; //將評價函數值最小節點置於隊頭}Close*shift(Open*q){ returnq->Array[--q->length];}//地圖初始化操作voidinitClose(Closecls[Height][Width],intsx,intsy,intdx,intdy){ //地圖Close表初始化配置 inti,j; for(i=0;i<Height;i++) { for(j=0;j<Width;j++) { cls[i][j].cur=&graph[i][j]; //Close表所指節點 cls[i][j].vis=!graph[i][j].reachable; //是否被訪問 cls[i][j].from=NULL; //所來節點 cls[i][j].G=cls[i][j].F=0; cls[i][j].H=abs(dx-i)+abs(dy-j); //評價函數值 } } cls[sx][sy].F=cls[sx][sy].H; //起始點評價初始值 // cls[sy][sy].G=0; //移步花費代價值 cls[dx][dy].G=Infinity;}voidinitGraph(constintmap[Height][Width],intsx,intsy,intdx,intdy){ //地圖發生變化時重新構造地 inti,j; srcX=sx; //起點X坐標 srcY=sy; //起點Y坐標 dstX=dx; //終點X坐標 dstY=dy; //終點Y坐標 for(i=0;i<Height;i++) { for(j=0;j<Width;j++) { graph[i][j].x=i;//地圖坐標X graph[i][j].y=j;//地圖坐標Y graph[i][j].value=map[i][j]; graph[i][j].reachable=(graph[i][j].value==Reachable); //節點可到達性 graph[i][j].sur=0;//鄰接節點個數 if(!graph[i][j].reachable) { continue; } if(j>0) { if(graph[i][j-1].reachable) //left節點可以到達 { graph[i][j].sur|=West; graph[i][j-1].sur|=East; } if(i>0) { if(graph[i-1][j-1].reachable &&graph[i-1][j].reachable &&graph[i][j-1].reachable) //up-left節點可以到達 { graph[i][j].sur|=North_West; graph[i-1][j-1].sur|=South_East; } } } if(i>0) { if(graph[i-1][j].reachable) //up節點可以到達 { graph[i][j].sur|=North; graph[i-1][j].sur|=South; } if(j<Width-1) { if(graph[i-1][j+1].reachable &&graph[i-1][j].reachable &&map[i][j+1]==Reachable)//up-right節點可以到達 { graph[i][j].sur|=North_East; graph[i-1][j+1].sur|=South_West; } } } } }}intbfs(){ inttimes=0; inti,curX,curY,surX,surY; unsignedcharf=0,r=1; Close*p; Close*q[MaxLength]={&close[srcX][srcY]}; initClose(close,srcX,srcY,dstX,dstY); close[srcX][srcY].vis=1; while(r!=f) { p=q[f]; f=(f+1)%MaxLength; curX=p->cur->x; curY=p->cur->y; for(i=0;i<8;i++) { if(!(p->cur->sur&(1<<i))) { continue; } surX=curX+dir[i].x; surY=curY+dir[i].y; if(!close[surX][surY].vis) { close[surX][surY].from=p; close[surX][surY].vis=1; close[surX][surY].G=p->G+1; q[r]=&close[surX][surY]; r=(r+1)%MaxLength; } } times++; } returntimes;}intastar(){ //A*演算法遍歷 //inttimes=0; inti,curX,curY,surX,surY; floatsurG; Openq;//Open表 Close*p; initOpen(&q); initClose(close,srcX,srcY,dstX,dstY); close[srcX][srcY].vis=1; push(&q,close,srcX,srcY,0); while(q.length) { //times++; p=shift(&q); curX=p->cur->x; curY=p->cur->y; if(!p->H) { returnSequential; } for(i=0;i<8;i++) { if(!(p->cur->sur&(1<<i))) { continue; } surX=curX+dir[i].x; surY=curY+dir[i].y; if(!close[surX][surY].vis) { close[surX][surY].vis=1; close[surX][surY].from=p; surG=p->G+sqrt((curX-surX)*(curX-surX)+(curY-surY)*(curY-surY)); push(&q,close,surX,surY,surG); } } } //printf("times:%d ",times); returnNoSolution;//無結果}constintmap[Height][Width]={ {0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,1,1}, {0,0,1,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1}, {0,0,0,0,0,0,1,0,0,0,0,0,0,1,1,0,0,0,0,1}, {0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,1}, {0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0}, {0,0,0,1,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0}, {0,1,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0}, {0,0,0,0,1,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,0}, {0,1,0,0,0,0,1,0,0,0,0,0,0,1,0,1,0,0,0,1}, {0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0}};constcharSymbol[5][3]={"□","▓","▽","☆","◎"};voidprintMap(){ inti,j; for(i=0;i<Height;i++) { for(j=0;j<Width;j++) { printf("%s",Symbol[graph[i][j].value]); } puts(""); } puts("");}Close*getShortest(){ //獲取最短路徑 intresult=astar(); Close*p,*t,*q=NULL; switch(result) { caseSequential: //順序最近 p=&(close[dstX][dstY]); while(p) //轉置路徑 { t=p->from; p->from=q; q=p; p=t; } close[srcX][srcY].from=q->from; return&(close[srcX][srcY]); caseNoSolution: returnNULL; } returnNULL;}staticClose*start;staticintshortestep;intprintShortest(){ Close*p; intstep=0; p=getShortest(); start=p; if(!p) { return0; } else { while(p->from) { graph[p->cur->x][p->cur->y].value=Pass; printf("（%d，%d）→ ",p->cur->x,p->cur->y); p=p->from; step++; } printf("（%d，%d） ",p->cur->x,p->cur->y); graph[srcX][srcY].value=Source; graph[dstX][dstY].value=Destination; returnstep; }}voidclearMap(){ //ClearMapMarksofSteps Close*p=start; while(p) { graph[p->cur->x][p->cur->y].value=Reachable; p=p->from; } graph[srcX][srcY].value=map[srcX][srcY]; graph[dstX][dstY].value=map[dstX][dstY];}voidprintDepth(){ inti,j; for(i=0;i<Height;i++) { for(j=0;j<Width;j++) { if(map[i][j]) { printf("%s",Symbol[graph[i][j].value]); } else { printf("%2.0lf",close[i][j].G); } } puts(""); } puts("");}voidprintSur(){ inti,j; for(i=0;i<Height;i++) { for(j=0;j<Width;j++) { printf("%02x",graph[i][j].sur); } puts(""); } puts("");}voidprintH(){ inti,j; for(i=0;i<Height;i++) { for(j=0;j<Width;j++) { printf("%02d",close[i][j].H); } puts(""); } puts("");}intmain(intargc,constchar**argv){ initGraph(map,0,0,0,0); printMap(); while(scanf("%d%d%d%d",&srcX,&srcY,&dstX,&dstY)!=EOF) { if(within(srcX,srcY)&&within(dstX,dstY)) { if(shortestep=printShortest()) { printf("從（%d，%d）到（%d，%d）的最短步數是:%d ", srcX,srcY,dstX,dstY,shortestep); printMap(); clearMap(); bfs(); //printDepth(); puts((shortestep==close[dstX][dstY].G)?"正確":"錯誤"); clearMap(); } else { printf("從（%d，%d）不可到達（%d，%d） ", srcX,srcY,dstX,dstY); } } else { puts("輸入錯誤！"); } } return(0);}

C. 求8數碼A或A*演算法（用C語言）

題目地址:http://acm.pku.e.cn/JudgeOnline/problem?id=1077

BFS:

#include <iostream>
using namespace std;
int fac[10]={1,1};
bool tflag[9];
struct bbit{
unsigned int val:4;
};
struct bbbit
{
unsigned int val:2;
};
struct Node
{
bbit s[9],pos;
int step;
bbbit path[21],tag;
int hashval()
{
int ret=0,i,j,tmp;
memset(tflag,false,sizeof(tflag));
for(i=0;i<8;i++)
{
tmp=0;
for(j=0;j<s[i].val;j++)
if(!tflag[j])
tmp++;
ret+=tmp*fac[8-i];
tflag[s[i].val]=true;
}
return ret;
}
bool up()
{
if(pos.val<=2)return false;
s[pos.val].val^=s[pos.val-3].val;
s[pos.val-3].val^=s[pos.val].val;
s[pos.val].val^=s[pos.val-3].val;
path[step].val=0;
pos.val-=3;
return true;
}
bool down()
{
if(pos.val>=6)return false;
s[pos.val].val^=s[pos.val+3].val;
s[pos.val+3].val^=s[pos.val].val;
s[pos.val].val^=s[pos.val+3].val;
path[step].val=1;
pos.val+=3;
return true;
}
bool left()
{
if(pos.val==0||pos.val==3||pos.val==6)return false;
s[pos.val].val^=s[pos.val-1].val;
s[pos.val-1].val^=s[pos.val].val;
s[pos.val].val^=s[pos.val-1].val;
path[step].val=2;
pos.val--;
return true;
}
bool right()
{
if(pos.val==2||pos.val==5||pos.val==8)return false;
s[pos.val].val^=s[pos.val+1].val;
s[pos.val+1].val^=s[pos.val].val;
s[pos.val].val^=s[pos.val+1].val;
path[step].val=3;
pos.val++;
return true;
}
bool operator==(const Node&x)const
{
int i;
for(i=0;i<9;i++)if(s[i].val!=x.s[i].val)return false;
return true;
}
}Q[362880],S,A,tmp,top;
struct Hash
{
bool d1,d2;
Node D;
}hash[362880];
inline void mkfac(){int i;for(i=2;i<=9;i++)fac[i]=fac[i-1]*i;}
inline int eval(char c){return c=='x'?0:c-'0';}
void o(Node x,Node y)
{
int i;
for(i=1;i<=x.step;i++)
{
switch(x.path[i].val)
{
case 0:putchar('u');break;
case 1:putchar('d');break;
case 2:putchar('l');break;
case 3:putchar('r');break;
}
}
for(i=y.step;i>=1;i--)
switch(y.path[i].val){
case 0:putchar('d');break;
case 1:putchar('u');break;
case 2:putchar('r');break;
case 3:putchar('l');break;
}
puts("");
}
int main()
{
char buf[11];
int i,t,l,r;
bool flag;
mkfac();
while(NULL!=gets(buf))
{
t=0;
for(i=0;i<=7;i++)A.s[i].val=i+1;A.s[8].val=0;A.pos.val=8;
for(i=0;buf[i];i++)
{
if(buf[i]==' ')continue;
S.s[t].val=eval(buf[i]);
if(S.s[t].val==0)
S.pos.val=t;
t++;
}
l=r=0;
flag=false;
for(i=0;i<362880;i++)hash[i].d1=hash[i].d2=false;
S.step=0;S.tag.val=1;
A.step=0;A.tag.val=2;
Q[r++]=S;//tag.val:1
Q[r++]=A;//tag.val:2
while(l<=r)
{
top=Q[l++];
top.step++;
tmp=top;
if(tmp.up())
{
if(tmp.tag.val==1)
{
if(!hash[t=tmp.hashval()].d1)
{
hash[t].d1=true;
Q[r++]=tmp;
if(hash[t].d2&&hash[t].D==tmp)
{
//find ans...
o(tmp,hash[t].D);
goto AA;
}
if(!hash[t].d2)hash[t].D=tmp;
}
}
else
{
if(!hash[t=tmp.hashval()].d2)
{
hash[t].d2=true;
Q[r++]=tmp;
if(hash[t].d1&&hash[t].D==tmp)
{
//find ans...
o(hash[t].D,tmp);
goto AA;
}
if(!hash[t].d1)hash[t].D=tmp;
}
}
}

tmp=top;
if(tmp.down())
{
if(tmp.tag.val==1)
{
if(!hash[t=tmp.hashval()].d1)
{
hash[t].d1=true;
Q[r++]=tmp;
if(hash[t].d2&&hash[t].D==tmp)
{
//find ans...
o(tmp,hash[t].D);
goto AA;
}
if(!hash[t].d2)hash[t].D=tmp;
}
}
else
{
if(!hash[t=tmp.hashval()].d2)
{
hash[t].d2=true;
Q[r++]=tmp;
if(hash[t].d1&&hash[t].D==tmp)
{
//find ans...
o(hash[t].D,tmp);
goto AA;
}
if(!hash[t].d1)hash[t].D=tmp;
}
}
}

tmp=top;
if(tmp.left())
{
if(tmp.tag.val==1)
{
if(!hash[t=tmp.hashval()].d1)
{
hash[t].d1=true;
Q[r++]=tmp;
if(hash[t].d2&&hash[t].D==tmp)
{
//find ans...
o(tmp,hash[t].D);
goto AA;
}
if(!hash[t].d2)hash[t].D=tmp;
}
}
else
{
if(!hash[t=tmp.hashval()].d2)
{
hash[t].d2=true;
Q[r++]=tmp;
if(hash[t].d1&&hash[t].D==tmp)
{
//find ans...
o(hash[t].D,tmp);
goto AA;
}
if(!hash[t].d1)hash[t].D=tmp;
}
}
}
tmp=top;
if(tmp.right())
{
if(tmp.tag.val==1)
{
if(!hash[t=tmp.hashval()].d1)
{
hash[t].d1=true;
Q[r++]=tmp;
if(hash[t].d2&&hash[t].D==tmp)
{
//find ans...
o(tmp,hash[t].D);
goto AA;
}
if(!hash[t].d2)hash[t].D=tmp;
}
}
else
{
if(!hash[t=tmp.hashval()].d2)
{
hash[t].d2=true;
Q[r++]=tmp;
if(hash[t].d1&&hash[t].D==tmp)
{
//find ans...
o(hash[t].D,tmp);
goto AA;
}
if(!hash[t].d1)hash[t].D=tmp;
}
}
}
}
AA:flag=true;
if(!flag)
puts("unsolvable");
}
return 0;
}

A*:

#include <iostream>
#include <queue>
using namespace std;
int fac[10]={1,1};
struct Node
{
int s[9],step,pos;
char path[501];
int hashval()
{
int ret=0,i,j,tmp;
bool flag[9];
memset(flag,false,sizeof(flag));
for(i=0;i<8;i++)
{
tmp=0;
for(j=0;j<s[i];j++)
if(!flag[j])
tmp++;
ret+=tmp*fac[8-i];
flag[s[i]]=true;
}
return ret;
}
bool up()
{
if(pos<=2)return false;
s[pos]^=s[pos-3];
s[pos-3]^=s[pos];
s[pos]^=s[pos-3];
path[step]='u';
pos-=3;
return true;
}
bool down()
{
if(pos>=6)return false;
s[pos]^=s[pos+3];
s[pos+3]^=s[pos];
s[pos]^=s[pos+3];
path[step]='d';
pos+=3;
return true;
}
bool left()
{
if(pos==0||pos==3||pos==6)return false;
s[pos]^=s[pos-1];
s[pos-1]^=s[pos];
s[pos]^=s[pos-1];
path[step]='l';
pos--;
return true;
}
bool right()
{
if(pos==2||pos==5||pos==8)return false;
s[pos]^=s[pos+1];
s[pos+1]^=s[pos];
s[pos]^=s[pos+1];
path[step]='r';
pos++;
return true;
}
bool operator==(const Node&x)const
{
int i;
for(i=0;i<9;i++)if(s[i]!=x.s[i])return false;
return true;
}
void show()
{
int i,j;
for(i=0;i<=6;i+=3,cout<<endl)
for(j=i;j<=i+2;j++)
cout<<s[j];
}
bool operator<(const Node&x)const
{
int la=0,lb=0,i;
for(i=0;i<8;i++)if(s[i]!=i+1)la++;la+=(s[8]!=0);
for(i=0;i<8;i++)if(x.s[i]!=i+1)lb++;lb+=(x.s[8]!=0);
return la>lb;
}
}S,A,tmp,top;
priority_queue<Node> Q;
bool hash[362880];
void mkfac(){int i;for(i=2;i<=9;i++)fac[i]=fac[i-1]*i;}
int eval(char c){return c=='x'?0:c-'0';}
void output(Node x)
{
int i;
for(i=1;i<=x.step;i++)
putchar(x.path[i]);
puts("");
}
int main()
{
char buf[11];
int i,t,l,r;
bool flag;
mkfac();
while(NULL!=gets(buf))
{
t=0;
for(i=0;i<=7;i++)A.s[i]=i+1;A.s[8]=0;A.pos=8;
for(i=0;buf[i];i++)
{
if(buf[i]==' ')continue;
S.s[t]=eval(buf[i]);
if(S.s[t]==0)
S.pos=t;
t++;
}
l=r=0;
flag=false;
memset(hash,false,sizeof(hash));
S.step=0;
while(!Q.empty())Q.pop();
Q.push(S);
while(!Q.empty())
{
top=Q.top();
Q.pop();
top.step++;
tmp=top;
if(tmp.up())
{
if(!hash[t=tmp.hashval()])
{
hash[t]=true;
Q.push(tmp);
if(tmp==A)
{
//find ans...
output(tmp);
goto AA;
}
}
}
tmp=top;
if(tmp.down())
{
if(!hash[t=tmp.hashval()])
{
hash[t]=true;
Q.push(tmp);
if(tmp==A)
{
//find ans...
output(tmp);
goto AA;
}
}
}
tmp=top;
if(tmp.left())
{
if(!hash[t=tmp.hashval()])
{
hash[t]=true;
Q.push(tmp);
if(tmp==A)
{
//find ans...
output(tmp);
goto AA;
}
}
}
tmp=top;
if(tmp.right())
{
if(!hash[t=tmp.hashval()])
{
hash[t]=true;
Q.push(tmp);
if(tmp==A)
{
//find ans...
output(tmp);
goto AA;
}
}
}
}
AA:flag=true;
if(!flag)
puts("unsolvable");
}
return 0;
}

D. 求A* 演算法C語言源程序

#include <string.h>
#include <stdio.h>
#include <malloc.h>
#include <time.h>
#define NULL 0

const int nmax = 200;
const int nend = 99; /*終點坐標的代表點*/
static char achar[10][10];
static int npayo = 0; /*0 表示空 1為非空*/
static int npayc = 0; /*0 表示空 1為非空*/
static int npay_x = 0; /*起點*/
static int npay_y = 0;
static int nend_x = 9; /*終點*/
static int nend_y = 9;
static int nnewpay_x;
static int nnewpay_y;
static int ndian = 101;
static int nh;
static long i = 10000000L;

struct Spaydian
{
int ng;
int nf; /*路徑評分*/
int nmy_x; /*自己位置*/
int nmy_y;
int nfatherx; /*父節點*/
int nfathery;
int nflag; /* 0 wei O; 1 wei @ */
};
static struct Spaydian spaydian[200];

/* open close list 表 */
typedef struct spaylist
{
int n_f;
int n_x;
int n_y;
int nfather_x;
int nfather_y;
struct spaylist *next;
};
static struct spaylist *open_list, *close_list;

static void smline();
static int sjudge(int nx,int ny,int i); /*判斷在第nx列ny行向第i個方向走是否可以,可以返回1否則返回0。
i=1表示向右，2表示向下，3表示向左，4表示向上*/
static void spath();
static void spayshow(int nxx,int nyy);
static int sifopen( int nx,int ny); /*判斷點是否在 open 表上*/
static int sifclose(int nx,int ny); /*判斷點是否在 close 表上*/
static int snewx(int nx,int i);
static int snewy(int ny,int i);
static spaylist *creat(); /*建立鏈表*/
static spaylist *del(spaylist *head,int num_x,int num_y); /*刪除鏈表的結點*/
static spaylist *snewdianx(spaylist *head);/*新的點*/
static spaylist *snewdiany(spaylist *head);
static spaylist *insert(spaylist *head,int ndian); /* 點插入鏈表 */
static spaylist *srebirth(spaylist *head,int ndian); /*更新表*/

int main()
{
FILE *fp ;
char ach ;
int ni = 0 ; /*統計個數*/
int nii = 0; /*achar[nii][njj]*/
int njj = 0;
if ((fp=fopen("route.txt","rt")) == NULL) /* 判斷打開文件是否為空 */
{
printf("文件為空！~\n");
return 0;
/* exit(1);*/
}
ach = fgetc(fp);
while(ach != EOF)
{
if(ach == 'O' || ach == '@') /*當值為@或O時候*/
{
spaydian[ni].ng = 0;
spaydian[ni].nf = nmax;
spaydian[ni].nmy_x = njj;
spaydian[ni].nmy_y = nii;
spaydian[ni].nfathery = -1;
spaydian[ni].nfatherx = -1;
if(ach == '@')
{
spaydian[ni].nflag = 1;
}
else if(ach == 'O')
{
spaydian[ni].nflag = 0;
}
ni++;
achar[nii][njj] = ach;
njj++;
if(njj == 10)
{
nii++;
njj = 0;
}
} /*end if*/
ach = fgetc(fp);
}/*end while*/
smline(); /* a*演算法 */
fp=fopen("answer.txt","w");
for(int i=0;i<10;i++ )
{ for(int j=0;j<10;j++ )
{
printf("%c",achar[i][j]);
if(j==9)
printf("\n");
fprintf(fp,"%c",achar[i][j]);
if (j==9)
fprintf(fp,"\n");
}
}
fclose(fp);
return 0;
}

/* a* 演算法 */
static void smline()
{ close_list = open_list = NULL;
open_list = creat();
while(open_list != NULL) /* 當open 表不為空時 */
{
open_list = del(open_list,npay_x,npay_y); /*刪除open 鏈表的結點*/
if(npay_x == 9 && npay_y == 9)
{

achar[9][9] = '=';
spath(); /*尋找並畫出路徑*/
break;
}
for (int i=1; i<=4; i++) /*四個方向逐個行走，i=1向右 2向下 3向左 4向上*/
{
if (sjudge(npay_x,npay_y,i))
{

nnewpay_x = snewx(npay_x,i);
nnewpay_y = snewy(npay_y,i);
if(open_list != NULL)
npayo = sifopen(nnewpay_x,nnewpay_y) ; /*判斷點是否在 open 表中*/
else
npayo = 0;

if(close_list != NULL)
npayc = sifclose(nnewpay_x,nnewpay_y) ; /*判斷點是否在 close 表中*/
else
npayc = 0;
ndian = 10*nnewpay_x+nnewpay_y ;

if (npayo == 0 && npayc == 0 ) /*點不在open表也不在close表中*/
{
spaydian[ndian].ng = spaydian[10*npay_x+npay_y].ng + 1; /*更新點的基本信息*/
nh = (nend - ndian)/10 + (nend - ndian)%10 ;
spaydian[ndian].nf = spaydian[ndian].ng+nh;
spaydian[ndian].nfathery = npay_y;
spaydian[ndian].nfatherx = npay_x;
spaydian[ndian].nmy_y = nnewpay_y;
spaydian[ndian].nmy_x = nnewpay_x;

open_list = insert(open_list,ndian);/*點插入open 表中*/
}
else if (npayo == 1) /*點在open表中*/
{
spaydian[ndian].ng = spaydian[10*npay_x+npay_y].ng + 1;
nh = (nend - ndian)/10 + (nend - ndian)%10 ;
if(spaydian[ndian].nf > (spaydian[ndian].ng+nh) && spaydian[ndian].nf != nmax)
{
spaydian[ndian].nf = spaydian[ndian].ng+nh;
open_list = srebirth(open_list,ndian); /*點插入open 表中*/
}
}
else if(npayc == 1) /*新生成的點在close表中*/
{
spaydian[ndian].ng = spaydian[10*npay_x+npay_y].ng + 1;
nh = (nend - ndian)/10 + (nend - ndian)%10 ;
if(spaydian[ndian].nf > (spaydian[ndian].ng+nh) && spaydian[ndian].nf != nmax)
{
spaydian[ndian].nf = spaydian[ndian].ng+nh;
close_list = srebirth(close_list,ndian);
close_list = del(close_list,nnewpay_x,nnewpay_y); /*刪除close鏈表的結點*/
open_list = insert(open_list,ndian);/*點插入open 表中*/
}
}/*end else if*/
}/*end if*/
}/*end for*/
close_list = insert(close_list,(10*npay_x+npay_y));/*點插入close 表中*/
if(open_list != NULL)
{
npay_x = open_list->n_x;
npay_y = open_list->n_y;
}

}/*end while*/
if(open_list == NULL)
{printf("無路可走 \n");}
}

/*建立鏈表*/
spaylist *creat(void)
{
spaylist *head;
spaylist *p1;
int n=0;
p1=(spaylist*)malloc(sizeof(spaylist));
p1->n_f = 18;
p1->n_x = 0;
p1->n_y = 0;
p1->nfather_x = -1;
p1->nfather_x = -1;
p1->next = NULL;
head = NULL;
head=p1;
return(head);
}

/*刪除結點*/
spaylist *del(spaylist *head,int num_x,int num_y)
{
spaylist *p1, *p2;
if(head == NULL)
{
printf("\nlist null!\n");
return (head);
}
p1 = head;
while((num_y != p1->n_y ||num_x != p1->n_x )&& p1->next != NULL)
{
p2=p1;
p1=p1->next;
}
if(num_x == p1->n_x && num_y == p1->n_y )
{
if(p1==head)
head=p1->next;
else
p2->next=p1->next;
}

return (head);
}

/*輸出*/
static void spath()
{
int nxx;
int nyy;
nxx = spaydian[nend].nfatherx;
nyy = spaydian[nend].nfathery;

spayshow(nxx,nyy) ;
}

/*遞歸*/
void spayshow(int nxx,int nyy)
{ achar[nxx][nyy] = '=';
if( nxx != 0 || nyy != 0 )
{
int nxxyy = 10*nxx+nyy;
nxx = spaydian[nxxyy].nfatherx;
nyy = spaydian[nxxyy].nfathery;
spayshow(nxx,nyy);
}
}

/* 判斷周圍四個點是否可行 */
static int sjudge(int nx,int ny,int i)
{
if (i==1) /*判斷向右可否行走*/
{
if (achar[nx][ny+1]=='O' && ny<9)
{
return 1;
}
else
return 0;
}
else if (i==2) /*判斷向下可否行走*/
{
if (achar[nx+1][ny]=='O' && nx<9)
{
return 1;
}
else
return 0;
}
else if (i==3)/*判斷向左可否行走 */
{
if (ny > 0&&achar[nx][ny-1]=='O')
{
return 1;
}
else
return 0;
}
else if (i==4)/*判斷向上可否行走 */
{
if (nx > 0&&achar[nx-1][ny]=='O')
{
return 1;
}
else
return 0;
}
else
return 0;
}

/* 求新的x點 */
static int snewx(int nx,int i)
{
if(i == 1)
nx = nx;
else if(i == 2)
nx = nx+1;
else if(i == 3)
nx = nx;
else if(i == 4)
nx = nx-1;
return nx;
}
/* 求新的y點 */
static int snewy(int ny, int i)
{
if(i == 1)
ny = ny+1;
else if(i == 2)
ny = ny;
else if(i == 3)
ny = ny-1;
else if(i == 4)
ny = ny;
return ny;
}

/*判定點是否在open表中*/
int sifopen(int nx,int ny)
{
spaylist *p1, *p2;
p1 = open_list;
while((ny != p1->n_y || nx != p1->n_x )&& p1->next != NULL)
{
p2 = p1;
p1 = p1->next;
}
if(nx == p1->n_x && ny == p1->n_y)
return 1;
else
return 0;
}

/*判定點是否在close表中*/
int sifclose(int nx,int ny)
{

spaylist *p1, *p2;
p1 = close_list;
while((ny != p1->n_y ||nx != p1->n_x )&& p1->next != NULL)
{
p2=p1;
p1=p1->next;
}
if(nx == p1->n_x && ny == p1->n_y)
return 1;
else
return 0;
}

/*插入結點*/
spaylist * insert(spaylist *head,int ndian)
{
spaylist *p0,*p1,*p2;
p1=head;
p0=(spaylist*)malloc(sizeof(spaylist));
p0->n_f = spaydian[ndian].nf;
p0->n_x = spaydian[ndian].nmy_x;
p0->n_y = spaydian[ndian].nmy_y;
p0->nfather_x = spaydian[ndian].nfatherx;
p0->nfather_x = spaydian[ndian].nfathery;
p0->next = NULL;
if(head==NULL)
{
head=p0;
p0->next=NULL;
}
else
{
while((p0->n_f > p1->n_f)&&(p1->next!=NULL))
{
p2=p1;
p1=p1->next;
}
if(p0->n_f <= p1->n_f)
{
if(head==p1)
head=p0;
else
p2->next=p0;
p0->next=p1;
}
else
{
p1->next=p0;
p0->next=NULL;
}
}
return (head);
}

/* 更新鏈表 */
spaylist * srebirth(spaylist *head,int ndian)
{
spaylist *p1, *p2;
p1=head;
while(spaydian[ndian].nmy_x!=p1->n_x&&spaydian[ndian].nmy_x!=p1->n_x&&p1->next!=NULL)
{
p2=p1;
p1=p1->next;
}
if(spaydian[ndian].nmy_x==p1->n_x&&spaydian[ndian].nmy_x==p1->n_x)
{
p1->n_f = spaydian[ndian].nf;
}
return (head);
}

E. 求一個A*演算法的C語言或C++代碼，小弟不勝感激，謝謝

1#include <iostream>
2#include <queue>
3usingnamespace std;
4
5struct knight{
6int x,y,step;
7int g,h,f;
8booloperator< (const knight & k) const{ //重載比較運算符
9return f > k.f;
10 }
11}k;
12bool visited[8][8]; //已訪問標記(關閉列表)
13int x1,y1,x2,y2,ans; //起點(x1,y1),終點(x2,y2),最少移動次數ans
14int dirs[8][2]={{-2,-1},{-2,1},{2,-1},{2,1},{-1,-2},{-1,2},{1,-2},{1,2}};//8個移動方向
15priority_queue<knight> que; //最小優先順序隊列(開啟列表)
16
17boolin(const knight & a){ //判斷knight是否在棋盤內
18if(a.x<0|| a.y<0|| a.x>=8|| a.y>=8)
19returnfalse;
20returntrue;
21}
22int Heuristic(const knight &a){ //manhattan估價函數
23return (abs(a.x-x2)+abs(a.y-y2))*10;
24}
25void Astar(){ //A*演算法
26 knight t,s;
27while(!que.empty()){
28 t=que.top(),que.pop(),visited[t.x][t.y]=true;
29if(t.x==x2 && t.y==y2){
30 ans=t.step;
31break;
32 }
33for(int i=0;i<8;i++){
34 s.x=t.x+dirs[i][0],s.y=t.y+dirs[i][1];
35if(in(s) &&!visited[s.x][s.y]){
36 s.g = t.g +23; //23表示根號5乘以10再取其ceil
37 s.h = Heuristic(s);
38 s.f = s.g + s.h;
39 s.step = t.step +1;
40 que.push(s);
41 }
42 }
43 }
44}
45int main(){
46char line[5];
47while(gets(line)){
48 x1=line[0]-'a',y1=line[1]-'1',x2=line[3]-'a',y2=line[4]-'1';
49 memset(visited,false,sizeof(visited));
50 k.x=x1,k.y=y1,k.g=k.step=0,k.h=Heuristic(k),k.f=k.g+k.h;
51while(!que.empty()) que.pop();
52 que.push(k);
53 Astar();
54 printf("To get from %c%c to %c%c takes %d knight moves.\n",line[0],line[1],line[3],line[4],ans);
55 }
56return0;
57}
58

F. 用C語言怎麼編一個a^n(a的n次方)的演算法

如果n比較小，可以吧
result
*=
a循環n次。。
如果n比較大，
可以逐步來算。
這樣考慮，f(n)
=
2^n
如果有了
f(m)的結果，
那麼
f(2m)和f(2m+1)
就分別等於
f(m)*f(m)和f(m)*f(m)*a
所以可以從最高位開始查看n的每一位，
如果這一位是1，
那麼
result
=
result
*
result
*
a;
如果這一位是0，那麼result
=
result
*
result;
其中result
的初始值是1。
這樣復雜度就是log(n)的

G. c語言中a+=a-=a*=a這個表達式的演算法是怎麼算的

a的初值呢？
a初值為12時，a+=a-=a*=a結果為0
步驟：
這個表達式的運算是從右向左的：
1.
a*=a：a=a*a=12*12=144
2.
a-=144:
a=a-144=144-144=0
3.
a+=0:
a=a+0=0+0=0