传送门HDU2612

描述

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.

输入

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

输出

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.

样例

Input

123456789101112131415164 4Y.#@.....#..@..M4 4Y.#@.....#..@#.M5 5Y..@..#....#...@..M.#...#

Output668866

题解

题意：地图上有多个KFC，小Y和小M一起从家出发前往一家KFC碰面，每单位可以往上下左右走，求在哪个KFC碰面所需时间最短，输出最短时间 分别以小Y和小M为起点跑两次BFS，预处理处理他们到所有KFC所需时间，然后遍历找最小和即可。

Code

1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859#include<bits/stdc++.h>#define INIT(a,b) memset(a,b,sizeof(a))#define LL long longusing namespace std;const int inf=0x3f3f3f3f;const int N=2e2+7;const int mod=1e9+7;char mp[N][N];int vis[N][N];int ans[N][N][2];int n,m,sx,sy,ex,ey;int rec[4][2]={1,0, -1,0, 0,1, 0,-1};struct node{ int i,j,time;};bool judge(int x,int y){ if(x<0||x>=n||y<0||y>=m||vis[x][y]||mp[x][y]=='#')return false; return true;}void bfs(int x,int y,int t){ INIT(vis,0); queue<node> que; que.push((node){x,y,0}); vis[x][y]=1; while(!que.empty()){ node q=que.front();que.pop(); if(mp[q.i][q.j]=='@') ans[q.i][q.j][t]=q.time * 11; for(int k=0;k<4;k++){ int tx=q.i+rec[k][0],ty=q.j+rec[k][1]; if(judge(tx,ty)){ vis[tx][ty]=1; que.push((node){tx,ty,q.time+1}); } } }}int main(){ while(cin>>n>>m){ INIT(ans,inf); getchar(); for(int i=0;i<n;i++){ for(int j=0;j<m;j++){ cin>>mp[i][j]; if(mp[i][j]=='Y') sx=i,sy=j; if(mp[i][j]=='M') ex=i,ey=j; } } bfs(sx,sy,0); bfs(ex,ey,1); int res=inf; for(int i=0;i<n;i++){ for(int j=0;j<m;j++){ res=min(ans[i][j][0]+ans[i][j][1],res); } } cout<<res<<endl; } return 0;}