题目

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

题解

有两个起点的BFS，每个KFS是一个出口，分别计算出每个出口距离两个入口的距离，计算其最小值

要特别注意：KFC有可能不能到达

代码

/*
By:OhYee
Github:OhYee
Email:oyohyee@oyohyee.com
Blog:http://www.cnblogs.com/ohyee/

かしこいかわいい？
エリーチカ！
要写出来Хорошо的代码哦~
*/

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
using namespace std;

//DEBUG MODE
#define debug 0

//循环
#define REP(n) for(int o=0;o<n;o++)

const int maxn = 205;
int n,m;
char Map[maxn][maxn];
const int delta[] = {1,-1,0,0};

struct point {
    int x,y;
    point() {
        x = y = -1;
    }
    point(int a,int b) {
        x = a;
        y = b;
    }
};
int num;


int BFS(point s,int (&dis)[maxn][maxn]) {
    memset(dis,-1,sizeof(dis));
    queue<point> Q;

    Q.push(s);
    dis[s.x][s.y] = 0;

    while (!Q.empty()) {
        int x = Q.front().x;
        int y = Q.front().y;
        Q.pop();

        REP(4) {
            int xx = x + delta[o];
            int yy = y + delta[3 - o];

            if (xx < 0 || xx >= n || yy < 0 || yy >= m)
                continue;
            if (Map[xx][yy] == '#')
                continue;
            if (dis[xx][yy] == -1) {
                dis[xx][yy] = dis[x][y] + 1;
                Q.push(point(xx,yy));
            }
        }
    }
    return -1;
}

bool Do() {
    if (scanf("%d%d",&n,&m) == EOF)
        return false;
    point s1,s2;
    num = 0;
    point v[maxn*maxn];
    for (int i = 0;i < n;i++)
        for (int j = 0;j < m;j++) {
            scanf("\n%c",&Map[i][j]);
            if (Map[i][j] == 'Y')
                s1 = point(i,j);
            if (Map[i][j] == 'M')
                s2 = point(i,j);
            if (Map[i][j] == '@') 
                v[num++] = point(i,j);
        }

    int dis1[maxn][maxn];
    int dis2[maxn][maxn];

    BFS(s1,dis1);
    BFS(s2,dis2);

    int Min = 100000;
    for (int i = 0;i < num;i++)
        if(dis1[v[i].x][v[i].y]!=-1&& dis2[v[i].x][v[i].y]!=-1)
        Min = min(Min,dis1[v[i].x][v[i].y]+ dis2[v[i].x][v[i].y]);

    printf("%d\n",Min * 11);

    return true;
}

int main() {
    while (Do());
    return 0;
}