hdu6667 Roundgod and Milk Tea（Hall定理）

题目链接

http://acm.hdu.edu.cn/showproblem.php?pid=6667

题目描述

Problem Description

Roundgod is a famous milk tea lover at Nanjing University second to none. This year, he plans to conduct a milk tea festival. There will be n classes participating in this festival, where the ith class has ai students and will make bi cups of milk tea.
Roundgod wants more students to savor milk tea, so he stipulates that every student can taste at most one cup of milk tea. Moreover, a student can't drink a cup of milk tea made by his class. The problem is, what is the maximum number of students who can drink milk tea?

Input

The first line of input consists of a single integer T (1≤T≤25), denoting the number of test cases.
Each test case starts with a line of a single integer n (1≤n≤10^6), the number of classes. For the next n lines, each containing two integers a,b (0≤a,b≤10^9), denoting the number of students of the class and the number of cups of milk tea made by this class, respectively.
It is guaranteed that the sum of n over all test cases does not exceed 6×10^6.

Output

For each test case, print the answer as a single integer in one line.

Sample Input

1

2

3 4

2 1

Sample Output

3

题意

有n个班级，每个班级有ai个人，可以做出bi杯奶茶，一个班级的人不能喝本班做的奶茶，可以喝别的班做的奶茶，问最多有多少人喝到奶茶。

思路

首先要了解Hall定理以及它的一个推论：

扫描二维码关注公众号，回复： 8846611 查看本文章

https://www.cnblogs.com/dummyummy/p/10311769.html

Hall定理我们离散数学学到过，想不到在这里用到了。。。

将人与奶茶分别看作二分图两边的顶点，班级则是所有顶点的子集，同一个班级的人与别的班级的奶茶之间有边相连，且任意两个不同班级的人可连接全部的奶茶。这样问题转化为了求二分图的最大匹配。利用Hall定理的推论很容易就能求出结果。

代码

#include<bits/stdc++.h>
#define ll long long
#define maxn 1000010
using namespace std;
ll a[maxn],b[maxn],n;

int main(){
	int t,i;
	cin>>t;
	while(t--){
		ll y=0,x=0,ans=0;
		scanf("%lld",&n);
		for(i=1;i<=n;i++){
			scanf("%lld%lld",&a[i],&b[i]);
			x+=a[i];
			y+=b[i];
		}
		if(x<y){
			for(i=1;i<=n;i++){
				ans=max(ans,a[i]-(y-b[i]));
			}
			printf("%lld\n",x-ans);
		}
		else{
			for(i=1;i<=n;i++){
				ans=max(ans,b[i]-(x-a[i]));
			}
			printf("%lld\n",y-ans);			
		}
	}
    return 0;
}