HDUOJ----（2612）Find a way

Find a way

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3177    Accepted Submission(s): 1031

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

简单的搜索，bfs，以Y和M为中心搜索

代码： 

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<deque>
using namespace std;
const int maxn=201;
typedef struct Map
{
  char ss;
  int x,y;
  int value;
}gmap;
unsigned int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};   /*·½Ïò*/
gmap bb[maxn][maxn];
int ans[maxn][maxn];
int n,m;
void bfs1(gmap a[][201],int x,int y)
{
    deque<gmap>q;
    gmap temp;
    q.push_back(a[x][y]);
    while(!q.empty())
    {
        temp=q.front();
        q.pop_front();
        for(int i=0;i<4/*&&temp.ss!='@'*/;i++)
        {

          if((temp.x+dir[i][0])<0||(temp.x+dir[i][0])>=n||(temp.y+dir[i][1])<0||(temp.y+dir[i][1])>=m)  continue;
          if(a[temp.x+dir[i][0]][temp.y+dir[i][1]].ss=='.'||a[temp.x+dir[i][0]][temp.y+dir[i][1]].ss=='@')
          {
                    a[temp.x+dir[i][0]][temp.y+dir[i][1]].value+=temp.value+1;
                  q.push_back(a[temp.x+dir[i][0]][temp.y+dir[i][1]]);
               if(a[temp.x+dir[i][0]][temp.y+dir[i][1]].ss=='@')
               {
                   a[temp.x+dir[i][0]][temp.y+dir[i][1]].ss='d';
                   ans[temp.x+dir[i][0]][temp.y+dir[i][1]]+=a[temp.x+dir[i][0]][temp.y+dir[i][1]].value;
               }
               else
                   a[temp.x+dir[i][0]][temp.y+dir[i][1]].ss='r';
          }
        }
    }
}

int main()
{
    int i,j,sx1,sy1,sx2,sy2; 
    while(scanf("%d%d",&n,&m)!=EOF)
    {
         getchar();
        memset(ans,0,sizeof(ans));
        for(i=0;i<n;i++)
        {
            for(j=0;j<m;j++)
            {
                scanf("%c",&bb[i][j].ss);
                bb[i][j].value=0;
                bb[i][j].x=i;
                bb[i][j].y=j;
                if(bb[i][j].ss=='Y')
                {
                    sx1=i;
                    sy1=j;
                }
                else if(bb[i][j].ss=='M')
                {
                    sx2=i;
                    sy2=j;
                }
            }
            getchar();
        }
        bfs1(bb,sx1,sy1);
        for(i=0;i<n;i++)
        {
              for(j=0;j<m;j++)
              {
                      bb[i][j].value=0;
                    if(bb[i][j].ss=='d')
                        bb[i][j].ss='@';
                    else
                     if(bb[i][j].ss=='r')
                        bb[i][j].ss='.';
              }
        }
        bfs1(bb,sx2,sy2);
       int     minm=INT_MAX;
            for(i=0;i<n;i++)
            {
              for(j=0;j<m;j++)
              {
                 if(bb[i][j].ss=='d'&&minm>ans[i][j])
                      minm=ans[i][j];
              }
            }
            printf("%d
",minm*11);
    }
    return 0;
}