poj 1693 二分图

 

非常水的二分图。。 不多说了 最近调一下模板 保证没有错误的 争取更快一点

 

#include <iostream>
#include <stdio.h>
#include <cstring>
using namespace std;

const int MAXN = 1002;
// 二分图匹配，返回最大的匹配数
// linker中保存匹配的关系 点从0开始
int g[MAXN][MAXN];
int n , m;
int linker[MAXN];
bool used[MAXN];

bool DFS ( int  a )
{
    for ( int i = 0 ; i < m; i ++ )
        if (g[a][i] && !used[i] )
        {
            used[i] = true ;
            if ( linker[i] == -1 || DFS( linker[i] ))
            {
                linker[i] = a;
                return true;
            }
        }
    return false;
}

int hungray ( )
{
    int result = 0;
    memset ( linker , -1 , sizeof ( linker )) ;
    for ( int i = 0 ; i < n; i ++ )
    {
        memset ( used , 0 , sizeof ( used )) ;
        if ( DFS ( i ) )
            result ++;
    }
    return result;
}

int main()
{
    int Case;
    scanf("%d" , &Case );
    while ( Case -- )
    {
        memset ( g, 0 , sizeof ( g ));
        int s;
        int a[7];
        int d , w;
        int sum = 0;
        int maxw = 0;
        scanf("%d" , &s );
        for ( int k = 0 ; k < s ;k ++ )
        {
            for ( int i = 0 ; i < 7 ; i ++ )
            {
                scanf("%d" , &a[i] );
            }
            scanf("%d %d", &d , &w );
            if ( w > maxw )
                maxw = w;
            for ( int i = sum ; i < sum+d ; i ++ )
            {
                for ( int j = 0 ; j < w ; j ++ )
                    for ( int t = 0 ; t < 7 ; t ++ )
                        if ( a[t] )
                            g[i][j*7+t] = 1;
            }
            sum += d;
        }
        n = sum;
        m = maxw*7;
        //cout << "sum == " << sum << endl;
        //cout << hungray()  << endl;
        if ( hungray() == sum )
            printf("Yes\n");
        else
            printf("No\n");

    }
    return 0;
}

// poj3041 把图看做成邻接矩阵就可以了 然后求二分图的最小点覆盖
// poj1325 将除0以外的点 构图 二分图的最小点覆盖
//