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 ",ans*11); 120 121 122 } 123 return 0; 124 }