zoj 1091 广搜。

属于比较水的广搜。。第一个是纯C 自己写的队列。。。第二个C++ STL的队列。。

#include <stdio.h>#include <string.h>#define MAXN 100000int xx[]={-2,-1,-2,-1,1,2,+1,+2};int yy[]={-1,-2,+1,+2,2,1,-2,-1};char tmp1[3],tmp2[3];int count;int visit[10][10];int i;struct lab{int x;int y;int step;};struct lab f,t;struct lab q[MAXN];struct lab tmp,tmp_q;int front=0,rear=0;int result;void push(struct lab k){ q[rear++] = k; }void pop(){front++;}void init(){f.x=tmp1[0]-'a'+1; f.y = tmp1[1]-'0';t.x=tmp2[0]-'a'+1; t.y = tmp2[1]-'0';memset(visit,0,sizeof(visit));front = rear = 0;}void bfs(int x,int y){tmp.x = x;tmp.y = y;tmp.step = 0;push(tmp);while (front<=rear){tmp = q[front];/*printf("into%d/n",tmp.step);printf("%d %d/n",tmp.x,tmp.y);*/pop();tmp_q = tmp;if(tmp.x==t.x&&tmp.y==t.y){/*printf("success/n");*/result = tmp.step;return ;}for(i=0;i<8;i++){tmp.x=tmp_q.x+xx[i];tmp.y=tmp_q.y+yy[i];tmp.step=tmp_q.step+1;if(tmp.x<9&&tmp.x>0&&tmp.y<9&&tmp.y>0&&!visit[tmp.x][tmp.y]){visit[tmp.x][tmp.y]=1;push(tmp);}}}}int main(void){while (scanf("%s %s",tmp1,tmp2)!=EOF){init();/*printf("%d %d %d %d/n",f.x,f.y,t.x,t.y);*/bfs(f.x,f.y);printf("To get from %s to %s takes %d knight moves./n",tmp1,tmp2,result);}return 0;}

 

 

STL版

#include <stdio.h>#include <string.h>#include <iostream>#include <queue>using namespace std;int xx[]={-2,-1,-2,-1,1,2,+1,+2};int yy[]={-1,-2,+1,+2,2,1,-2,-1};int visit[10][10];struct lab{int x;int y;int step;}p[2],tmp,tmp1;int result;queue <lab> q;void bfs(char x, int y){tmp.x = x;tmp.y = y;tmp.step = 0;q.push(tmp);while(!q.empty()){//printf("fds/n");tmp = q.front();q.pop();tmp1.x=tmp.x;tmp1.y=tmp.y;tmp1.step =tmp.step;if(tmp.x==p[1].x&&tmp.y==p[1].y){result = tmp1.step;//printf("result=%d/n",result);return ;}for(int i=0;i<8;i++){tmp.x = tmp1.x + xx[i];tmp.y = tmp1.y + yy[i];tmp.step = tmp1.step+1;if(tmp.x<9&&tmp.x>0&&tmp.y<9&&tmp.y>0&&!visit[tmp.x][tmp.y]){visit[tmp.x][tmp.y]=1;q.push(tmp);//printf("step=%d/n",tmp.step);}}}/*if(x<'a'||x>'h'||y<1||y>8)return ;*//*if(x==p[1].x&&y==p[1].y){printf("%d/n",count);return ;}*/}int main(void){//int h_f,h_t,x_f,x_t; //from to 横竖。 上下左右//char h_t,char waste,tmp11,tmp22;while(scanf("%c%d%c%c%d",&tmp11,&p[0].y,&waste,&tmp22,&p[1].y)!=EOF){p[0].x = tmp11-'a'+1; p[1].x = tmp22-'a'+1;//printf("%d/n",result);memset(visit,0,sizeof(visit));while(!q.empty())q.pop();//int front=0,rear=0;//printf("%d/n",result);bfs(p[0].x,p[0].y);//To get from e2 to e4 takes 2 knight moves.printf("To get from %c%d to %c%d takes %d knight moves./n",tmp11,p[0].y,tmp22,p[1].y,result);//printf("%d/n",result);//printf("...%d %d %d %d/n",p[0].x,p[0].y,p[1].x,p[1].y);getchar();}return 0;}