[bfs]HDU2612 Find a way

N - Find a way
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
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Status
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

搜索题，此题最大的创新之处在于每个点都可以搜到两段不同的终点，对其中每个kfc都做一次广度搜索，搜索出Y或M到同一个@点的最短路，总共有两组解，输出其中的最小值即可。

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <queue>#define N 210#define inf 0x3f3f3f3fusing namespace std;int m,n,mark[N][N],dis[N][N][2],dir[4][2]={1,0, 0,1, -1,0, 0,-1},flag;char s[N][N];struct node{   int x,y,step;};bool judge(int x,int y){   if(x>=0 && x<m && y>=0 && y<n && s[x][y]!='#' && mark[x][y]==0)       return 1;   return 0;}void bfs(int x,int y){   int k;   queue<node>q;   node cur,next;   cur.x=x;cur.y=y;cur.step=0;   mark[x][y]=1;   q.push(cur);   while(!q.empty())   {       cur=q.front();       q.pop();       next.step=cur.step+1;       for(k=0;k<4;k++)       {           next.x=x=cur.x+dir[k][0];           next.y=y=cur.y+dir[k][1];           if(judge(x,y))           {               mark[x][y]=1;               if(s[x][y]=='@')                   dis[x][y][flag]=next.step;               q.push(next);           }       }   }}int main(){  //  freopen("in.txt","r",stdin);   int i,j,min;   while(scanf("%d %d",&m,&n)!=-1)   {       min=inf;       for(i=0;i<m;i++)           for(j=0;j<n;j++)               dis[i][j][0]=dis[i][j][1]=inf;       for(i=0;i<m;i++)           scanf("%s",s[i]);       for(i=0;i<m;i++)           for(j=0;j<n;j++)           {               if(s[i][j]=='Y')               {                   flag=0;                   memset(mark,0,sizeof(mark));                   bfs(i,j);               }               else if(s[i][j]=='M')               {                   flag=1;                   memset(mark,0,sizeof(mark));                   bfs(i,j);               }           }       for(i=0;i<m;i++)           for(j=0;j<n;j++)               if(s[i][j]=='@' && min>dis[i][j][0]+dis[i][j][1])                   min=dis[i][j][0]+dis[i][j][1];           printf("%d\n",min*11);   }   return 0;}