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Find a wayTime Limit: 3000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6362Accepted Submission(s): 2116
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 4Y.#@…..#..@..M4 4Y.#@…..#..@#.M5 5Y..@..#….#…@..M.#…#
Sample Output
668866
Author
yifenfei有一点需要说明dis开了个三维数组第三维表示不同的两次搜索，每次给第三维传递一个参数来表示哪一次。另外对于dis初始化的时候应初始化无穷大，因为有可能某个KFC根本就无法到达，影响最后求结果。说说题外话：留校不回家真是够人……训练效果也突出不出来……#include<iostream>#include<sstream>#include<algorithm>#include<cstdio>#include<string.h>#include<cctype>#include<string>#include<cmath>#include<vector>#include<stack>#include<queue>#include<map>#include<set>using namespace std;const int INF=204;const int M=0x1f1f1f1f;int n,m,z;char cnt[INF][INF];int dir[][2]= {{1,0},{0,-1},{-1,0},{0,1}};int dis[INF][INF][2];int vis[INF][INF];struct Node{int x,y,step;Node(int x,int y,int z):x(x),y(y),step(z) {}Node() {}};void bfs(int x,int y){memset(vis,0,sizeof(vis));vis[x][y]=1;queue<Node>q;q.push(Node(x,y,0));while(!q.empty()){Node u=q.front();q.pop();for(int i=0; i<4; i++){int tx=u.x+dir[i][0];int ty=u.y+dir[i][1];if(!vis[tx][ty]&& tx>=1&&tx<=n&&ty>=1&&ty<=m&&cnt[tx][ty]!='#'){vis[tx][ty]=1;dis[tx][ty][z]=u.step+1;q.push(Node(tx,ty,u.step+1));}}}}int main(){while(cin>>n>>m){for(int i=0;i<2;i++)for(int j=1;j<=n;j++)for(int k=1;k<=m;k++)dis[j][k][i]=M;for(int i=1; i<=n; i++)scanf("%s",cnt[i]+1);for(int i=1; i<=n; i++)for(int j=1; j<=m; j++){if(cnt[i][j]=='Y'){z=1;bfs(i,j);}if(cnt[i][j]=='M'){z=0;bfs(i,j);}}int ans=M;for(int i=1; i<=n; i++){for(int j=1; j<=m; j++){if(cnt[i][j]=='@'){ans=min(ans,dis[i][j][0]+dis[i][j][1]);}}}cout<<ans*11<<endl;}}
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