BFS-HDU-2612-Find a Way

Find a way

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6615 Accepted Submission(s): 2189

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

分别从两个人脚下出发进行BFS，求出到每个KFC的最短距离，最后再求一个最小的距离和就行了。

//
// main.cpp
// 简单搜索-N-Find a Way
//
// Created by 袁子涵 on 15/8/17.
// Copyright (c) 2015年 袁子涵. All rights reserved.
//

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <stdlib.h>
#define minof(a,b) ((a)>(b)?(b):(a))
using namespace std;

typedef struct dt
{
    int x,y;
    long int step;
}Data;

typedef struct queue
{
    long int head,tail;
    Data data[50000];
}Queue;

Queue q;

int n,m;
char map[210][210];
bool visit1[210][210];
bool visit2[210][210];
long int step1[210][210];
long int step2[210][210];
int dir[2][4]={-1,1,0,0,0,0,-1,1};
unsigned long long int sum;

void bfs(int i,int j,bool visit[210][210],long int step[210][210])
{
    q.head=0;
    q.tail=0;
    dt t,k;
    t.x=i;
    t.y=j;
    t.step=0;
    q.data[q.tail++]=t;
    while (q.head!=q.tail) {
        t=q.data[q.head++];
        for (int i=0; i<4; i++) {
            k=t;
            k.x+=dir[0][i];
            k.y+=dir[1][i];
            k.step++;
            if (k.x>=0 && k.x<n && k.y>=0 && k.y<m && visit[k.x][k.y]==0 && map[k.x][k.y]!='#') {
                step[k.x][k.y]=k.step;
                visit[k.x][k.y]=1;
                q.data[q.tail++]=k;
            }
        }
    }
}

int main(int argc, const char * argv[]) {
    while (scanf("%d%d",&n,&m)!=EOF) {
        memset(map, 0, sizeof(map));
        memset(visit1, 0, sizeof(visit1));
        memset(visit2, 0, sizeof(visit2));
        memset(step1, 0, sizeof(step1));
        memset(step2, 0, sizeof(step2));
        sum=40000;
        for (int i=0; i<n; i++) {
            for (int j=0; j<m; j++) {
                cin >> map[i][j];
            }
        }
        for (int i=0; i<n; i++) {
            for (int j=0; j<m ; j++) {
                if (map[i][j]=='Y') {
                    visit1[i][j]=1;
                    step1[i][j]=0;
                    bfs(i,j,visit1,step1);
                }
                if (map[i][j]=='M') {
                    visit2[i][j]=1;
                    step2[i][j]=0;
                    bfs(i,j,visit2,step2);
                }
            }
        }
        for (int i=0; i<n; i++) {
            for (int j=0; j<m; j++) {
                if (map[i][j]=='@' && visit1[i][j] && visit2[i][j]) {
                    sum=minof(step1[i][j]+step2[i][j],sum);
                }
            }
        }
        sum*=11;
        cout << sum << endl;
    }
    return 0;
}