HDU2612(Find a way)搜索-两次BFS

/*****************************************************题目大意：Y和M要在KFC约会,地图上有多个KFC,求在哪个KFC两人所走总距离最短;算法思想：两个人到同一点的和最短,分别以两个人为起点,BFS遍历整个图;计算出到每个KFC的最短距离,然后枚举所有的KFC,求出最小的代价即可;******************************************************/#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<queue>#include<algorithm>using namespace std;const int N=205;const int MAX=0xffffff;int x1,y1,x2,y2;//两个人的起点int v1[N][N];//计算路程和标记是否访问int v2[N][N];//计算路程和标记是否访问char map[N][N];//地图int dir[4][2]= {{1,0},{-1,0},{0,1},{0,-1}};//上下左右int n,m;struct node//坐标集{    int x;    int y;};void BFS(int x ,int y,int v[][N]){    queue<node>Q;    node p;    p.x=x;    p.y=y;    Q.push(p);    v[x][y]=0;    while(!Q.empty())    {        node p1=Q.front();        Q.pop();        for(int i=0; i<4; i++)        {            node p2;            p2.x=p1.x+dir[i][0];            p2.y=p1.y+dir[i][1];            if(v[p2.x][p2.y]==MAX&&p2.x>=0&&p2.x<n&&p2.y>=0&&p2.y<m&&map[p2.x][p2.y]!='#')            {                v[p2.x][p2.y]=v[p1.x][p1.y]+11;                Q.push(p2);            }        }    }}void Solve(){    for(int i=0; i<n; i++)    {        for(int j=0; j<m; j++)        {            v1[i][j]=v2[i][j]=MAX;        }    }    BFS(x1,y1,v1);    BFS(x2,y2,v2);    int Min=MAX;    for(int i=0; i<n; i++)    {        for(int j=0; j<m; j++)        {            if(map[i][j]=='@'&&v1[i][j]!=MAX&&v2[i][j]!=MAX)//枚举所有的KFC            {                if(Min>v1[i][j]+v2[i][j])                    Min=v1[i][j]+v2[i][j];            }        }    }    printf("%d\n",Min);}int main(){    freopen("C:\\Users\\Administrator\\Desktop\\kd.txt","r",stdin);    while(~scanf("%d%d",&n,&m))    {        char ch;        for(int i=0; i<n; i++)        {            for(int j=0; j<m; j++)            {                scanf(" %c",&ch);                map[i][j]=ch;                if(ch=='Y')                {                    x1=i;                    y1=j;                }                if(ch=='M')                {                    x2=i;                    y2=j;                }            }        }        Solve();    }    return 0;}