hdu 2612 Find a way（bfs）

Problem Description

Pass a year learning in Hangzhou, yifenfei arrival hometown Ningbo at finally. Leave Ningbo one year, yifenfei have many people to meet. Especially a good friend Merceki.
Yifenfei’s home is at the countryside, but Merceki’s home is in the center of city. So yifenfei made arrangements with Merceki to meet at a KFC. There are many KFC in Ningbo, they want to choose one that let the total time to it be most smallest. 
Now give you a Ningbo map, Both yifenfei and Merceki can move up, down ,left, right to the adjacent road by cost 11 minutes.

Input

The input contains multiple test cases.
Each test case include, first two integers n, m. (2<=n,m<=200). 
Next n lines, each line included m character.
‘Y’ express yifenfei initial position.
‘M’    express Merceki initial position.
‘#’ forbid road;
‘.’ Road.
‘@’ KCF

Output

For each test case output the minimum total time that both yifenfei and Merceki to arrival one of KFC.You may sure there is always have a KFC that can let them meet.

Sample Input

4 4
Y.#@
....
.#..
@..M
4 4
Y.#@
....
.#..
@#.M
5 5
Y..@.
.#...
.#...
@..M.
#...#

Sample Output

66
88
66

bfs求出两个人到各个点的距离，最后枚举最短的距离。用c++提交WA，G++提交AC了，这是什么原因。。。

  1 #pragma comment(linker, "/STACK:1024000000,1024000000")
  2 #include<iostream>
  3 #include<cstdio>
  4 #include<cstring>
  5 #include<cmath>
  6 #include<math.h>
  7 #include<algorithm>
  8 #include<queue>
  9 #include<set>
 10 #include<bitset>
 11 #include<map>
 12 #include<vector>
 13 #include<stdlib.h>
 14 #include <stack>
 15 using namespace std;
 16 #define PI acos(-1.0)
 17 #define max(a,b) (a) > (b) ? (a) : (b)
 18 #define min(a,b) (a) < (b) ? (a) : (b)
 19 #define ll long long
 20 #define eps 1e-10
 21 #define MOD 1000000007
 22 #define N 206
 23 #define inf 1e12
 24 int n,m;
 25 char mp[N][N];
 26 struct Node{
 27    int x,y;
 28    int t;
 29 }st1,st2;
 30 int vis[N][N];
 31 int dirx[]={0,0,-1,1};
 32 int diry[]={-1,1,0,0};
 33 int dist1[N][N];
 34 int dist2[N][N];
 35 void bfs1(Node st){
 36    memset(vis,0,sizeof(vis));
 37    queue<Node>q;
 38    q.push(st);
 39    vis[st.x][st.y]=1;
 40 
 41    Node t1,t2;
 42    while(!q.empty()){
 43       t1=q.front();
 44       q.pop();
 45       for(int i=0;i<4;i++){
 46          t2.x=t1.x+dirx[i];
 47          t2.y=t1.y+diry[i];
 48          if(mp[t2.x][t2.y]=='#') continue;
 49          if(t2.x<0 || t2.x>=n || t2.y<0 || t2.y>=m) continue;
 50          if(vis[t2.x][t2.y]) continue;
 51          vis[t2.x][t2.y]=1;
 52          t2.t=t1.t+1;
 53          dist1[t2.x][t2.y]=t2.t;
 54          q.push(t2);
 55       }
 56    }
 57 }
 58 void bfs2(Node st){
 59    memset(vis,0,sizeof(vis));
 60    queue<Node>q;
 61    q.push(st);
 62    vis[st.x][st.y]=1;
 63 
 64    Node t1,t2;
 65    while(!q.empty()){
 66       t1=q.front();
 67       q.pop();
 68       for(int i=0;i<4;i++){
 69          t2.x=t1.x+dirx[i];
 70          t2.y=t1.y+diry[i];
 71          if(mp[t2.x][t2.y]=='#') continue;
 72          if(t2.x<0 || t2.x>=n || t2.y<0 || t2.y>=m) continue;
 73          if(vis[t2.x][t2.y]) continue;
 74          vis[t2.x][t2.y]=1;
 75          t2.t=t1.t+1;
 76          dist2[t2.x][t2.y]=t2.t;
 77          q.push(t2);
 78       }
 79    }
 80 }
 81 int main()
 82 {
 83    while(scanf("%d%d",&n,&m)==2){
 84       memset(dist1,0,sizeof(dist1));
 85       memset(dist2,0,sizeof(dist2));
 86       for(int i=0;i<n;i++){
 87          scanf("%s",mp[i]);
 88          for(int j=0;j<m;j++){
 89             if(mp[i][j]=='Y'){
 90                st1.x=i;
 91                st1.y=j;
 92                st1.t=0;
 93             }
 94             if(mp[i][j]=='M'){
 95                st2.x=i;
 96                st2.y=j;
 97                st2.t=0;
 98             }
 99          }
100       }
101 
102       bfs1(st1);
103       bfs2(st2);
104 
105       int ans=inf;
106       for(int i=0;i<n;i++){
107          for(int j=0;j<m;j++){
108             if(mp[i][j]=='@'){
109                if(dist1[i][j]!=0 && dist2[i][j]!=0){
110                   int sum=dist1[i][j]+dist2[i][j];
111                   if(sum<ans){
112                      ans=sum;
113                   }
114                }
115 
116             }
117          }
118       }
119       printf("%d\n",ans*11);
120 
121 
122    }
123     return 0;
124 }