Find a way

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6221    Accepted Submission(s): 2070

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
[email protected]
.#...
.#...
@..M.
#...#

Sample Output

66
88
66

Author

yifenfei

Source

奋斗的年代

一开始写成了让对没个肯德基店做一次bfs,会TLE

之后发现然对两个人分别做一次bfs就行

只是遇到肯德基店的 时候,因为不能确定这个是否是做优的,所以不return,继续搜下去.

/*************************************************************************
    > File Name: code/2015summer/searching/N.cpp
    > Author: 111qqz
    > Email: [email protected]
    > Created Time: 2015年07月25日 星期六 13时54分49秒
 ************************************************************************/

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#define y0 abc111qqz
#define y1 hust111qqz
#define yn hez111qqz
#define j1 cute111qqz
#define tm crazy111qqz
#define lr dying111qqz
using namespace std;
#define REP(i, n) for (int i=0;i<int(n);++i)
typedef long long LL;
typedef unsigned long long ULL;
const int N= 2E2+5;
char st[N][N];
int n,m;
int d[N][N];
int dd[N][N];
int dx[4]={0,0,-1,1};
int dy[4]={1,-1,0,0};
struct node
{
    int x,y;
}kfc[N];
node M,Y;
void bfs(int xo,int yo)
{
    memset(d,-1,sizeof(d));
    queue<int>x;
    queue<int>y;
    x.push(xo);
    y.push(yo);
    d[xo][yo]=0;
    while (!x.empty())
    {
       int px=x.front();x.pop();
       int py=y.front();y.pop();
       for ( int i = 0 ; i < 4 ; i++ )
       {
         int nx = px+dx[i];
         int ny = py+dy[i];
         if (nx>=0&&nx<n&&ny>=0&&ny<m&&d[nx][ny]==-1&&st[nx][ny]!=‘#‘)
         {
              d[nx][ny]=d[px][py]+1;
             x.push(nx);
             y.push(ny);
         }
       }

    }
  //  cout<<"res1:"<<res<<endl;
}
void bfs2(int xo,int yo)
{
    memset(dd,-1,sizeof(dd));
    queue<int>x;
    queue<int>y;
    x.push(xo);
    y.push(yo);
    dd[xo][yo]=0;
    while (!x.empty())
    {
      int px = x.front();x.pop();
      int py =y.front();y.pop();
      for ( int i = 0 ;  i <4 ; i++ )
      {
        int nx = px +dx[i];
        int ny = py +dy[i];
        if (nx>=0&&nx<n&&ny>=0&&ny<m&&dd[nx][ny]==-1&&st[nx][ny]!=‘#‘)
        {
            dd[nx][ny]= dd[px][py]+1;
            x.push(nx);
            y.push(ny);
        }
      }
    }

}

int main()
{
    while (scanf("%d %d",&n,&m)!=EOF)
    {
      for ( int i = 0 ; i < n ; i++ )
      {
        cin>>st[i];
      }
      int k = 0;
      for ( int i = 0 ;  i < n ; i++)
      {
        for ( int j = 0 ; j < m ; j++ )
        {
            if (st[i][j]==‘Y‘)
            {
              Y.x=i;
               Y.y=j;
            }
            if (st[i][j]==‘M‘)
            {
               M.x=i;
              M.y=j;
             }
        }
      }
      bfs(Y.x,Y.y);
      bfs2(M.x,M.y);
      int ans = 99999999;
      for ( int i =  0 ; i < n ; i++ )
      {
        for ( int j = 0 ;  j < m ; j++ )
        {
            if (st[i][j]==‘@‘)
            {
             // cout<<d[i][j]<<"  "<<dd[i][j]<<endl;
              if (d[i][j]==-1||dd[i][j]==-1) continue;
              ans = min(ans,d[i][j]+dd[i][j]);
            }
        }
      }

      cout<<ans*11<<endl;
    }

    return 0;
}