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最新推荐文章于 2022-11-30 20:08:15 发布

Little_boy_z

最新推荐文章于 2022-11-30 20:08:15 发布

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本文链接：https://blog.youkuaiyun.com/Little_boy_z/article/details/51523578

本文介绍了一种通过广度优先搜索(BFS)算法确定两个人从不同起点出发到达多个指定会面地点中最近的一个的方法。考虑到地图上的障碍物及可达路径，通过计算每个人到达地图上每个点的时间，并选择总时间最短的会面地点。

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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


他不只有一个KFC！这意味着我们需要求出所有的来，找最小值。那样我们可以设立一个数组，记录人到x，y点处所用的时间，之后这要这个点是KFC，我们就更新min!!!!
除此之外就是一个很经典的BFS了
#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
#include<algorithm>
#include<stdio.h>
using namespace std;
struct node
{
    int x;
    int y;
    int count;
};
int n,m;
char ditu[1005][1005];
bool visit[1005][1005];
int to[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
int manx,many,womanx,womany;
int mantime[1005][1005],womentime[1005][1005];
deque<node>Q;
void BFS(node p,int mantime[][1005])
{
    int i;
    node q;
    memset(visit,0,sizeof(visit));
    memset(mantime,0,sizeof(mantime));
    Q.push_front(p);
    while(!Q.empty())
    {
        p=Q.front();
        Q.pop_front();
        for(i=0;i<4;i++)
        {
            q.x=p.x+to[i][0];
            q.y=p.y+to[i][1];
            q.count=p.count+1;
            if (q.x>=0&&q.y>=0&&q.x<m&&q.y<n&&ditu[q.x][q.y]!='#'&&!visit[q.x][q.y])
             {
                 visit[q.x][q.y]=1;
                 mantime[q.x][q.y]=q.count;
                 Q.push_back(q);
             }
        }
    }
}
int main()
{
    int i,j,min;
    node p;
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        for(i=0;i<m;i++)
            for(j=0;j<n;j++)
           {
               cin>>ditu[i][j];
               if(ditu[i][j]=='Y')
                 {
                  manx=i;
                  many=j;
                 }
                else if(ditu[i][j]=='M')
                {
                    womanx=i;
                    womany=j;
                }
           }
    }
    min=9999999;
    p.x=manx;
    p.y=many;
    p.count=0;
    BFS(p,mantime);
    p.x=womanx;
    p.y=womany;
    p.count=0;
    BFS(p,womentime);
     for (i=0;i<m;i++)
            for (j=0;j<n;j++)
                if (ditu[i][j]=='@')
                    if (mantime[i][j]!=0&&womentime[i][j]!=0)
                        if(mantime[i][j]+womentime[i][j]<min)/*比较求距离的最小值*/
                            min = mantime[i][j]+womentime[i][j];


        printf("%d/n",min*11);




}