HDU 2612 Find a way

http://acm.hdu.edu.cn/showproblem.php?pid=2612

Find a way

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4478    Accepted Submission(s): 1525

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
     
     
     
     

 

Author

yifenfei

 

Source

奋斗的年代

 

Recommend

yifenfei

两次BFS  先求Y到所有KFC 的最短距离 在求M到所有KFC的最短距离 加和 求最短的

暴力枚举KFC 求两次BFS 会超时  BFS次数为 n*2 (n为KFC数量)----不知道当时怎么想的、

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<vector>
#include<cmath>
#include<stdlib.h>
#include<iomanip>
#include<list>
#include<deque>
#include<map>
#include <stdio.h>
#include <queue>
#include <stack>
#define maxn 10000+5
#define ull unsigned long long
#define ll long long
#define reP(i,n) for(i=1;i<=n;i++)
#define rep(i,n) for(i=0;i<n;i++)
#define cle(a) memset(a,0,sizeof(a))
#define mod 90001
#define PI 3.141592657
#define INF 1<<30
const ull inf = 1LL << 61;
const double eps=1e-5;

using namespace std;
struct node
{
	int x,y;
	int dist;
};
bool cmp(int a,int b)
{
    return a>b;
}
int n,m;
char mp[210][210];
int vis[210][210];
int dir[4][2]={1,0,0,1,0,-1,-1,0};
int val[210][210];
queue<node>q;
void bfs(node a)
{
	while(!q.empty())q.pop();
	cle(vis);
	vis[a.x][a.y]=1;
	node u;
	u.dist=0;
	u.x=a.x;
	u.y=a.y;
	q.push(u);
	while(!q.empty())
	{
		u=q.front();q.pop();
		if(mp[u.x][u.y]=='@')val[u.x][u.y]+=u.dist;
		for(int i=0;i<4;i++)
		{
			node v;
			v.x=u.x+dir[i][0];
			v.y=u.y+dir[i][1];
			if(!vis[v.x][v.y]&&mp[v.x][v.y]!='#'&&v.x<=n&&v.x>=1&&v.y<=m&&v.y>=1)
			{
				v.dist=u.dist+1;
				vis[v.x][v.y]=1;
				q.push(v);
			}
		}
	}
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);
    while(scanf("%d%d",&n,&m)!=EOF)
	{
		cle(val);
		node Y,M;
		for(int i=1;i<=n;i++)
			for(int j=1;j<=m;j++)
			{
				cin>>mp[i][j];
				if(mp[i][j]=='M')M.x=i,M.y=j,M.dist=0;
				if(mp[i][j]=='Y')Y.x=i,Y.y=j,Y.dist=0;
			}
			int ans=INF;
			bfs(Y),bfs(M);
		for(int i=1;i<=n;i++)
			for(int j=1;j<=m;j++)
			{
				if(val[i][j]!=0)
				ans=min(ans,val[i][j]);
			}
		printf("%d\n",ans*11);
	}
    return 0;
}