bfs经典题，起点和终点分别bfs，借助于结构体，队列，枚举最后的共同终点所需的步数……
#include<bits/stdc++.h>
using namespace std;
char Map[205][205];
int vis[205][205],n,m,cnt;
int dir[4][2]={{0,1},{1,0},{-1,0},{0,-1}};
struct node
{
int x;
int y;
int step;
};
node path1[1001];
node path2[1001];
void bfs(int x,int y,node path[])
{
int i;
memset(vis,0,sizeof(vis));
queue<node>q;
node s,e;
s.x=x,s.y=y,s.step=0;
q.push(s);
while(!q.empty())
{
s=q.front();
q.pop();
if(Map[s.x][s.y]=='@') path[cnt++]=s;
for(int i=0;i<4;i++)
{
int newx=s.x+dir[i][0];
int newy=s.y+dir[i][1];
if(newx<0||newy<0||newx>=n||newy>=m) continue;
if(Map[newx][newy]=='#') continue;
if(vis[newx][newy]) continue;
vis[newx][newy]=1;
e.x=newx,e.y=newy,e.step=s.step+1;
q.push(e);
}
}
}
int main()
{
while(scanf("%d%d",&n,&m)==2)
{
memset(path1,0,sizeof(path1));
memset(path2,0,sizeof(path2));
cnt=0;
getchar();
int x1,y1,x2,y2;
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
scanf("%c",&Map[i][j]);
if(Map[i][j]=='Y') x1=i,y1=j;
if(Map[i][j]=='M') x2=i,y2=j;
}
getchar();
}
bfs(x1,y1,path1);
int cnt1=cnt;
cnt=0;
bfs(x2,y2,path2);
int cnt2=cnt;
int minn=INT_MAX;
for(int i=0;i<cnt1;i++)
{
for(int j=0;j<cnt2;j++)
{
if(path1[i].x==path2[j].x&&path1[i].y==path2[j].y)
minn=min(minn,11*(path1[i].step+path2[j].step));
}
}
printf("%d\n",minn);
}
return 0;
}