暑假小训练 （一道 不能不会的用结构体搜索题）

http://acm.hust.edu.cn/vjudge/contest/129524#problem/I

题意：8x8的象棋，给出开始和结束的点，和一个不能走的点，求出最短路。

#include<iostream>#include<cstdio>#include<cstring>#include<queue>using namespace std;int ax,ay,bx,by,cx,cy;int gragh[10][10];int path[10][10];class Map       //用结构体可以方便的判断x，y{public:    int x,y;}now;void bfs(){    now.x = ax;    now.y = ay;    queue<Map>q;    q.push(now);    while(!q.empty()){        Map  a = q.front();        q.pop();        if(a.x == bx && a.y == by)            return ;        for(int i = -1;i <= 1; i++)            for(int j = -1;j <= 1; j++)                if(i != 0 || j != 0){                    if(a.x+i <= 8 && a.x+i >= 1 && a.y+j <= 8 && a.y+j >= 1 && gragh[a.x+i][a.y+j] != -1 && path[a.x+i][a.y+j] == 0)                    {       //分别判断了：界，不是c点，没走过                        Map temp;                        temp.x = a.x+i;                        temp.y = a.y+j;                        path[temp.x][temp.y] = path[a.x][a.y]+1;                        q.push(temp);                    }                }    }}int main(){    int ncase = 1;//    freopen("in.txt","r",stdin);    while(scanf("%d%d%d%d%d%d",&ax,&ay,&bx,&by,&cx,&cy) != EOF){        memset(path,0,sizeof(path));        memset(gragh,0,sizeof(gragh));  //记得初始化        gragh[cx][cy] = -1;        bfs();        printf("Case %d: %d\n",ncase++,path[bx][by]);    }    return 0;}