Find a way hdu2612

Find a way

Time Limit : 3000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 3   Accepted Submission(s) : 3
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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

AC代码：
#include<stdio.h>
#include<memory.h>
int step[4][2]={{1,0},{0,1},{-1,0},{0,-1}};
char map[201][201];
int z[201][201],v[201][201],Y_x,Y_y,M_x,M_y,n,m;
struct none
{
    int x,y;
    int count;
}q[40001];
void BFS(int x,int y)
{
    int i;
    int fornt,end,tempx,tempy,count;
    memset(z,0,sizeof(z));
    fornt=1;end=1;
    q[end].x=x;
    q[end].y=y;
    q[end].count=0;
    end++;
    while(fornt!=end)
    {
        for(i=0;i<4;i++)
        {
            tempx=q[fornt].x+step[i][0];
            tempy=q[fornt].y+step[i][1];
            if(tempx>=0&&tempx<n&&tempy>=0&&tempy<m)
            {
                if(map[tempx][tempy]!='#'&&z[tempx][tempy]==0)
                {
                    q[end].x=tempx;
                    q[end].y=tempy;
                    q[end].count=q[fornt].count+1;
                    end++;
                    z[tempx][tempy]=1;
                    if(map[tempx][tempy]=='@')
                    {
                        v[tempx][tempy]+=q[end-1].count;
                    }
                }
            }
        }
        fornt++;
    }
}
int main()
{
    freopen("IO.txt","r",stdin);
    int i,j,min;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(i=0;i<n;i++)
            scanf("%s",map[i]);
        for(i=0;i<n;i++)
            for(j=0;j<m;j++)
            {
                if(map[i][j]=='Y')
                {
                    Y_x=i;
                    Y_y=j;
                }
                if(map[i][j]=='M')
                {
                    M_x=i;
                    M_y=j;
                }
            }
        min=999999999;
        memset(v,0,sizeof(v));
        BFS(Y_x,Y_y);
        BFS(M_x,M_y);
        for(i=0;i<n;i++)
            for(j=0;j<m;j++)
            {
                if(min>v[i][j]&&v[i][j]!=0)
                    min=v[i][j];
            }
        printf("%d/n",min*11);
    }
    return 0;
}

解题思路：
此题要用求出最短的路程，可以用广搜，因为有多个KFC，所以难以知道怎么才是最短的，可以先把Y去每个KFC的路程求出，存在V数组中，然后把M去每个KFC的路程求出，加到V数组中，这样只要遍历V就可以得出最短的路径了，两次广搜加遍历

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