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Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 13005 Accepted Submission(s): 3258

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

2 2

1 0

1 1

Sample Output

YES

AC代码

#include <bits/stdc++.h>

#define pb push_back

#define mp make_pair

#define fi first

#define se second

#define all(a) (a).begin(), (a).end()

#define fillchar(a, x) memset(a, x, sizeof(a))

#define huan printf("\n")

#define debug(a,b) cout<<a<<" "<<b<<" "<<endl

#define ffread(a) fastIO::read(a)

using namespace std;

typedef long long ll;

typedef pair<int,int> pii;

const int MAXN = 1e5+;//左集合大小

const int MAXM = ;//右集合大小

int uN,vN;//左右集合的数量

int g[MAXN][MAXM];

int linker[MAXM][MAXN];

bool used[MAXM];

int num[MAXM];//右集合的限制

bool dfs(int u)

{

    for(int v = ; v < vN; v++)

        if(g[u][v] && !used[v])

        {

            used[v] = true;

            if(linker[v][] < num[v])

            {

                linker[v][++linker[v][]] = u;

                return true;

            }

            for(int i = ; i <= num[v]; i++)

                if(dfs(linker[v][i]))

                {

                    linker[v][i] = u;

                    return true;

                }

        }

    return false;

}

int hungary()

{

    int res = ,flag=;

    for(int i = ; i < vN; i++)

        linker[i][] = ;

    for(int u = ; u < uN; u++)

    {

        memset(used,false,sizeof(used));

        if(dfs(u))

            res++;

        else   //当前人不能匹配 直接就是NO了

        {

            flag=;

            break;

        }

    }

    return flag;

}

int main()

{

    while(scanf("%d%d",&uN,&vN)!=EOF)

    {

        for(int i=;i<uN;i++)

            for(int j=;j<vN;j++)

                scanf("%d",&g[i][j]);

        for(int i=;i<vN;i++)

            scanf("%d",&num[i]);

        if(hungary())

            puts("NO");

        else

            puts("YES");

    }

}

巴特西

HDU 3609 二分图多重匹配

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