HDU2612 Find a way BFS

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本文介绍了一种解决两人从不同起点寻找最短时间总和到达同一目的地的问题。通过两次广度优先搜索(BFS)预处理每个人到达地图上每个点的最短步数，进而求得两人到达任一目标地点的时间总和最小值。

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Find a way

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 28250 Accepted Submission(s): 9128

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

问题连接

问题描述

有两个人要从他们各自的起点到地图上的KFC见面，每十一分钟他们可以从当前位置向地图的上下左右其中一个方向移动到下一个区域，其中’#'表示路障，不能到达。要求输出他们分别到KFC的时间之和中的最小值。如：他们到其中一家KFC用的时间分别为T1,T2，那么时间之和为T=T1+T2,要找到最小的T。

问题分析

使用两次BFS分别记录他们到达地图上的任意一点要用的最小的步数，其中就包括了到不同KFC的步数。再遍历一次地图，就可以找到最小的和，然后转化为时间输出就可以了。

程序如下

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
const int N = 205;

int n, m;

char map[N][N];//地图
int dir[4][2] = { { 0, 1 }, { 1, 0 }, { 0, -1 }, { -1, 0 } };//方向
int dist[2][N][N];//Y,M打表

struct state
{
    int y, x, ti;
}q, p;

void bfs(char, bool);//BFS
int find_min();//找出最小的
state find(char);

int main()
{
    while (scanf("%d %d", &n, &m) != EOF)
    {
        for (int i = 0; i < n; i++)//输入地图 
            scanf("%s", map[i]);
        bfs('Y', 0);//Y打表 
        bfs('M', 1);//M打表 
        int road = find_min();//找到两个人到KFC的最小的时间和 
        printf("%d\n", road * 11);
    }
    return 0;
}

void bfs(char name, bool flag)
{
    queue<state>Q;
    memset(dist[flag], 0x3f, sizeof(dist[flag]));
    q = find(name);//找到位置 
    Q.push(q);
    while (!Q.empty())
    {
        q = Q.front();
        Q.pop();
        for (int i = 0; i < 4; i++)
        {
            p = q;
            p.x += dir[i][0];
            p.y += dir[i][1];
            p.ti++;
            //要求到达这个位置时用时是最短的以及地图合理
            if (dist[flag][p.y][p.x]>p.ti&&p.x >= 0 && p.x < m&&p.y >= 0 && p.y < n&&map[p.y][p.x] != '#') 
            {
                dist[flag][p.y][p.x] = p.ti;
                Q.push(p);
            }
        }
    }
}
state find(char start)
{
    int i, j;
    for (i = 0; i < n; i++)
        for (j = 0; j < m; j++)
            if (map[i][j] == start)
            {
                state aim = { i, j, 0 };
                return aim;
            }
}
int find_min()
{
    int i, j, min;
    min = 0x3f3f3f3f;
    for (i = 0; i < n; i++)
        for (j = 0; j < m; j++)
            if (map[i][j] == '@'&&min>(dist[0][i][j] + dist[1][i][j]))
                min = dist[0][i][j] + dist[1][i][j];
    return min;
}