HDU 6667 Roundgod and Milk Tea (思维)

2019 杭电多校 8 1011

题目链接：HDU 6667

比赛链接：2019 Multi-University Training Contest 8

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 $a_i$ students and will make $b_i$ 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 $T (1\le T\le 25)$, denoting the number of test cases.

Each test case starts with a line of a single integer $n (1\le n\le 10^6)$, the number of classes. For the next $n$ lines, each containing two integers $a,b (0\le a,b\le 10^9)$, 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 $6\times 10^6$.

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

Solution

题意：

有 $n$ 个班级，每个班有 $a_i$ 个人，做了 $b_i$ 杯奶茶，每个班的每个人最多喝一杯奶茶且不能和自己班做的奶茶，问最多共有多少人喝到奶茶。

思路

最初的想法是用一个 $sum$ 记录所有剩余的奶茶数，然后每个组能喝的奶茶数为 $sum -$ 该组的奶茶(自己不能喝自己的) $+$ 上一组做的奶茶 (上一组减掉的加回来)。后来发现有点问题，就是中间一步减掉自己的奶茶可能是减多的，也就是上一组喝掉的可能就是当前组的奶茶，那么当前组剩余的奶茶是比原来少的，于是就用 $tmp2$ 保存上一组喝掉的奶茶数，每次让上一组喝掉当前组的奶茶，如果不够喝再用 $tmp$ 保存还要喝掉的奶茶数，往下迭代。

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

const int maxn = 1e6 + 10;

struct Team
{
    ll m, n;  // 人数 奶茶数
} t[maxn];

int cmp(Team t1, Team t2) {
    return t1.m > t2.m;
}

int main() {
    int T;
    cin >> T;
    while(T--) {
        int n;
        scanf("%d", &n);
        ll sum = 0;
        for(int i = 0; i < n; ++i) {
            scanf("%lld%lld", &t[i].m, &t[i].n);
            sum += t[i].n;
        }
        sort(t, t + n, cmp);
        ll ans = 0; 
        ll tmp = t[0].n; // tmp 保存喝掉的奶茶数 第一组一定要被喝
        ll tmp2 = 0; // tmp2 保存的是上一组喝掉的奶茶
        for(int i = 0; i < n; ++i) {
            if(i) {
                // 上一组喝掉的奶茶数+之前喝掉的奶茶数
                if(t[i].n < tmp2 + tmp) {
                    t[i].n = 0;
                    tmp = tmp2 + tmp - t[i].n;
                } else {
                    t[i].n = t[i].n - (tmp2 + tmp);
                    tmp = 0;
                }
            }
            sum -= t[i].n; // 自己不能喝自己的奶茶
            if(i) sum += t[i - 1].n; // 可以喝上一组的奶茶
            // 剩余的奶茶数与第 i 组人数比较
            if(sum >= t[i].m) {
                ans += t[i].m;
                sum -= t[i].m;
                tmp2 = t[i].m;
            } else {
                ans += sum;
                sum -= sum;
                tmp2 = sum;
            }
            // cout << ans << endl;
        }
        printf("%lld\n", ans);
    }
    return 0;
}

比赛中完全想复杂了，其实完全可以很快处理。把每个人能喝的奶茶加起来和所有的奶茶比较即可。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e6 + 10;

ll a[maxn], b[maxn];

int main() {
    int T;
    scanf("%d", &T);
    while(T--) {
        int n;
        scanf("%d", &n);
        ll sum = 0;
        for(int i = 0; i < n; ++i) {
            scanf("%lld%lld", &a[i], &b[i]);
            sum += b[i];
        }
        ll ans = 0;
        for(int i = 0; i < n; ++i) {
            ans += min(a[i], sum - b[i]);
        }
        printf("%lld\n", min(ans, sum));
    }
    return 0;
}