2019 Multi-University Training Contest 8 Roundgod and Milk Tea

最新推荐文章于 2020-07-11 19:23:30 发布

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本文探讨了一个有趣的问题：在南京大学的奶茶节上，如何最大化可以让学生品尝到奶茶的人数。通过精心设计的算法，我们解决了班级间奶茶分配的挑战，确保每个学生都能享受到一杯来自其他班级的手制奶茶。

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Roundgod and Milk Tea

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≤106), the number of classes. For the next n lines, each containing two integers a,b (0≤a,b≤10⁹), 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 6×106.

Output

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

Sample Input

Sample Output

代码

#include<bits/stdc++.h>
using namespace std;
typedef struct {
    long long a,b;
}node;
bool cmp(node a,node b){
    return a.b<b.b;
}
node m[1000006];
long long s[1000006];
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        int n;
        scanf("%d",&n);
        long long sumn=0,sum=0;;
        for(int i=0;i<n;i++){
            scanf("%d%d",&m[i].a,&m[i].b);
            sumn+=(long long)m[i].a;
        }
        for(int i=0;i<n;i++){
            s[i]=sumn-m[i].a;
        }
        for(int i=0;i<n;i++){
            sum+=min(s[i],m[i].b);
        }
        printf("%lld\n",min(sum,sumn));
    }
    return 0; 
}