HDU 3605 Escape (二分图的多重匹配)

最新推荐文章于 2020-09-09 17:47:25 发布

原创最新推荐文章于 2020-09-09 17:47:25 发布 · 533 阅读

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本文介绍了一种解决二分图中多重匹配问题的算法实现，该问题涉及确定一定数量的人是否可以全部适配到若干个星球上居住，每个星球有不同的容纳限制。文章通过具体的代码示例展示了如何通过深度优先搜索进行匹配。

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Escape

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2481 Accepted Submission(s): 686

Problem Description
2012 If this is the end of the world how to do? I do not know how. But now scientists have found that some stars, who can live, but some people do not fit to live some of the planet. Now scientists want your help, is to determine what all of people can live in these planets.

Input
More set of test data, the beginning of each data is n (1 <= n <= 100000), m (1 <= m <= 10) n indicate there n people on the earth, m representatives m planet, planet and people labels are from 0. Here are n lines, each line represents a suitable living conditions of people, each row has m digits, the ith digits is 1, said that a person is fit to live in the ith-planet, or is 0 for this person is not suitable for living in the ith planet.
The last line has m digits, the ith digit ai indicates the ith planet can contain ai people most..
0 <= ai <= 100000

Output
Determine whether all people can live up to these stars
If you can output YES, otherwise output NO.

Sample Input
1 1
1
1

2 2
1 0
1 0
1 1

Sample Output
YES
NO

/*
  二分图的多重匹配
  和最大匹配类似，只不过记录连接的边时，不再是只记录
  一个，而是记录点能容纳的匹配数量。
  
  703MS	8000K
  */
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define SIZE 100005

using namespace std;

int N,M,ans;
int limit[15];
bool cnt[SIZE][15],vis[15];
int link[15][SIZE],num[SIZE];
int temp[15];

bool dfs(int lt)
{
    for(int i=1; i<=M; i++)
    {
        if(cnt[lt][i] && !vis[i])
        {
            vis[i] = true;
            if(num[i] < limit[i])
            {
                link[i][++num[i]] = lt;
                return true;
            }
            for(int j=1; j<=num[i]; j++)
            {
                if(dfs(link[i][j]))
                {
                    link[i][j] = lt;
                    return true;
                }
            }
        }
    }
    return false;
}

int main()
{
    while(~scanf("%d%d",&N,&M))
    {
        ans = 0;
        int t;
        for(int i=1; i<=N; i++)
        {
            for(int j=1; j<=M; j++)
            {
                scanf("%d",&t);
                cnt[i][j] = t? true:false;
            }
        }
        for(int i=1; i<=M; i++)
            scanf("%d",&limit[i]);
        memset(link,0,sizeof(link));
        memset(num,0,sizeof(num));
        for(int i=1; i<=N; i++)
        {
            memset(vis,0,sizeof(vis));
            if(!dfs(i))
            {
                ans = 1;
                break;
            }
        }
        if(ans == 0) puts("YES");
        else puts("NO");
    }
    return 0;
}