HDU3605 Escape —— 二分图多重匹配

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3605

 

Escape

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 10349    Accepted Submission(s): 2476

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

 

题解：

二分图多重匹配的裸题，但要适当修改以防止超时。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
#define ms(a,b) memset((a),(b),sizeof((a)))
//#define LOCAL
using namespace std;
typedef long long LL;
const int INF = 2e9;
const LL LNF = 9e18;
const int mod = 1e9+7;
const int maxn = 1e5+10;

int n, m;
int linker[20][maxn];
int g[maxn][20];
bool used[20];
int num[20];

bool dfs(int u)
{
    for(int v = 1; v<=m; v++)
    if(g[u][v] && !used[v])
    {
        used[v] = true;
        if(linker[v][0] < num[v])
        {
            linker[v][++linker[v][0]] = u;
            return true;
        }
        for(int i = 1; i<=num[v]; i++)
        if(dfs(linker[v][i]))
        {
            linker[v][i] = u;
            return true;
        }
    }
    return false;
}

bool hungary()
{
    for(int v = 1; v<=m; v++)
        linker[v][0] = 0;
    for(int u = 1; u<=n; u++)
    {
        memset(used,0,sizeof(used));
        if(!dfs(u)) return false; //如果这个人匹配不到，则直接退出。否则会超时。
    }
    return true;
}

int main()
{
#ifdef LOCAL
    freopen("123", "r", stdin);
//      freopen("output.txt", "w", stdout);
#endif

    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(int i = 1; i<=n; i++)
        for(int j = 1; j<=m; j++)
            scanf("%d",&g[i][j]);
        for(int i = 1; i<=m; i++)
            scanf("%d",&num[i]);

        if(hungary()) puts("YES");
        else puts("NO");
    }
}