kuangbin专题一：N题，HDU2612:Find a way

HDU2612:Find a way

kuangbin专题一：N题

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 17428    Accepted Submission(s): 5605

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

 题意：两个人碰面吃饭 ， 每走一步用时11min 问两人耗费总时间最小是多少

思路：以两个人的位置为起点 ， 两次广度优先搜索 ， 由于存在多个饭馆 遇到一个饭馆就 存一下费用（耗时） ，

取 两人到同一位置的最小耗时和

/* 不想加注释 ......*/
#include <Cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std ; 

#define maxn 300 
bool visit[maxn][maxn] ; 
int n , m ; 
char map[maxn][maxn] ; 
int result1[maxn][maxn] ; 
int result2[maxn][maxn] ; 
int x1 , y1 , x2 , y2 ; 
int dirx[] = {0 , 0 , -1 , 1 } ; 
int diry[] = {-1 , 1 , 0 , 0 } ; 

struct node {
    int x ; 
    int y ; 
    int step ; 
};

bool check(int x , int y ){
    if(x>=1 && x<= n && y>=1 && y<=m){
        return true ; 
    }
    return false ; 
}

void BFS(int x , int y , int result[][maxn]){
    memset(visit , 0 , sizeof(visit)) ; 
    
    queue<node> Q ;  
    node st ; 
    st.x = x ; 
    st.y = y ; 
    st.step = 0 ; 
    Q.push(st) ; 
    while(!Q.empty()){
        node q = Q.front() ; 
        Q.pop() ; 
        if(map[q.x][q.y] == '@'){
            result[q.x][q.y] = q.step ; 
        }
        node turn ; 
        for(int i=0 ; i<4 ; i++){
            turn.x = q.x + dirx[i] ; 
            turn.y = q.y + diry[i] ; 
            turn.step = q.step + 1 ; 
            if(check(turn.x , turn.y ) && !visit[turn.x][turn.y] && map[turn.x][turn.y]!='#' ) {
                visit[turn.x][turn.y] = 1 ;
                Q.push(turn) ; 
            }
        }
        
    }
    return;
}

int main(){
    
    while(~scanf("%d%d" , &n , &m)){
        for(int i=1 ; i<=n ; i++){
            for(int j=1 ; j<=m ; j++){
                scanf(" %c" , &map[i][j]) ; 
                if(map[i][j] == 'Y'){
                    x1 = i ; 
                    y1 = j ; 
                }
                if(map[i][j] == 'M'){
                    x2 = i ; 
                    y2 = j ; 
                }
            }
        }
        
        memset(result1 , 0 , sizeof(result1)) ; 
        memset(result2 , 0 , sizeof(result2)) ;
        
        BFS(x1 , y1 , result1) ; 
        BFS(x2 , y2 , result2) ;
        
        int min_num = 1000000 ; 
        for(int i=1 ; i<=n ; i++){
            for(int j=1 ; j<= m ; j++){
                if(result1[i][j] == 0 || result2[i][j] == 0 || map[i][j] != '@')
                    continue ; 
                    
                min_num = min(result1[i][j] + result2[i][j] , min_num ) ; 
            }
        } 
        printf("%d
" , min_num * 11 ) ; 
    } 
    return 0 ; 
}