hdu 2612 Find a way(双向bfs)

最新推荐文章于 2021-07-22 10:23:03 发布

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#C++ #bfs

本文介绍了一个利用双向广度优先搜索(BFS)解决寻找两个角色从各自起点出发到同一目标点所需的最短时间的问题。通过给定的城市地图和初始位置，算法能够找出到达任一指定KFC所需总时间最小的方案。

Find a way

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4820 Accepted Submission(s): 1635

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

Author

yifenfei

题目分析:Y,M不能走,需要注意,然后就是裸的双向bfs了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#define MAX 207

using namespace std;

typedef pair<int,int> PII;

int n,m,yx,yy,mx,my;
char s[MAX];
int mp[MAX][MAX];
int mark[3][MAX][MAX];
vector<PII> v;

struct Node
{
    int x,y,t;
    Node ( int a, int b , int c )
        : x(a),y(b),t(c){}
};

int dx[]={1,0,-1,0};
int dy[]={0,1,0,-1};

void bfs ( int flag , int sx , int sy )
{
    queue<Node> q;
    q.push ( Node( sx , sy , 0 ) );
    mark[flag][sx][sy] = 0;
    while ( !q.empty() )
    {
        int x = q.front().x;
        int y = q.front().y;
        int t = q.front().t;
        q.pop();
        for ( int i = 0 ; i < 4 ; i++ )
        {
            int tx = x + dx[i] , ty = y + dy[i];
            if ( mp[tx][ty] == -1 ) continue;
            if ( mark[flag][tx][ty] != -1 ) continue;
            q.push( Node ( tx , ty , t+1 ) );
            mark[flag][tx][ty] = t+1;
        }
    }
}

int main ( )
{
    while ( ~scanf ( "%d%d" , &n , &m ) )
    {
        v.clear();
        memset ( mark , -1 , sizeof ( mark ) );
        memset ( mp , -1 , sizeof ( mp ) );
        for ( int i = 1 ; i <= n ; i++ )
        {
            scanf ( "%s" , s+1 );
            for ( int j = 1 ; j <= m ; j++ )
                if ( s[j] == 'Y' )
                    mp[i][j] = -1 , yx = i , yy = j;
                else if ( s[j] == '.' ) 
                    mp[i][j] = 1;
                else if ( s[j] == '@' )
                    mp[i][j] = 1,v.push_back(make_pair(i,j));
                else if ( s[j] == 'M' )
                    mp[i][j] = -1 , mx = i , my = j;
        }
        bfs ( 0 , yx , yy );
        bfs ( 1 , mx , my );
        int len = v.size();
        int ans = 1000000;
        for ( int i = 0 ; i < len ; i++ )
        {
            if ( mark[0][v[i].first][v[i].second] == -1 ) continue;
            if ( mark[1][v[i].first][v[i].second] == -1 ) continue;
            ans = min ( ans , mark[0][v[i].first][v[i].second]
                             +mark[1][v[i].first][v[i].second] );
        }
        printf ( "%d\n" , 11*ans );
    }
}