NC25025 [USACO 2007 Nov G]Sunscreen

题目

题目描述

To avoid unsightly burns while tanning, each of the $C$ ($1 ≤ C ≤ 2500$) cows must cover her hide with sunscreen when they're at the beach. Cow $i$ has a minimum and maximum $SPF$ rating ($1 ≤ minSPF_i ≤ 1,000; minSPF_i ≤ maxSPF_i ≤ 1,000$) that will work. If the $SPF$ rating is too low, the cow suffers sunburn; if the $SPF$ rating is too high, the cow doesn't tan at all........

The cows have a picnic basket with $L$ ($1 ≤ L ≤ 2500$) bottles of sunscreen lotion, each bottle $i$ with an $SPF$ rating $SPF_i$ ($1 ≤ SPFi ≤ 1,000$). Lotion bottle $i$ can cover $cover_i$ cows with lotion. A cow may lotion from only one bottle.

What is the maximum number of cows that can protect themselves while tanning given the available lotions?

输入描述

• Line $1$ : Two space-separated integers: $C$ and $L$
• Lines $2 \cdots C+1$ : Line $i$ describes cow $i$'s lotion requires with two integers: $minSPF_i$ and $maxSPF_i$
• Lines $C+2 \cdots C+L+1$ : Line $i+C+1$ describes a sunscreen lotion bottle $i$ with space-separated integers: $SPF_i$ and $cover_i$

输出描述

A single line with an integer that is the maximum number of cows that can be protected while tanning.

示例1

输入

3 2
3 10
2 5
1 5
6 2
4 1

输出

2

题解

知识点：枚举，贪心。

我们希望使用防晒的牛是最优选择，即在自己有独立机会时不占用别的牛的可用机会。因此，考虑建立一个能够选择机会少的序列能够优先选择的排序。对于一系列由同一同侧端点的区间，选择最靠近另一端点的防晒是最优的，因为这样使得对小区间的机会影响最小化。再者，通过对这个同一端点从小到大排序，对于每一组同一同侧端点如上操作，就能每次使得选择最优。

我们选择对区间右端点从小到大排序，因此要选择最靠左的可行防晒，于是防晒因从小到大排序来选择。对每右端点对应一系列区间，从小到大选择防晒，使得防晒最靠左。

时间复杂度 $O(cl)$

空间复杂度 $O(c+l)$

代码

#include <bits/stdc++.h>
 
using namespace std;
 
pair<int, int> a[2507], b[2507];
 
int main() {
    std::ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int c, l;
    cin >> c >> l;
    for (int i = 0;i < c;i++) cin >> a[i].first >> a[i].second;
    for (int i = 0;i < l;i++) cin >> b[i].first >> b[i].second;
    sort(a, a + c, [&](pair<int, int> a, pair<int, int> b) {return a.second < b.second;});
    sort(b, b + l);
    int cnt = 0;
    for (int i = 0;i < c;i++) {
        for (int j = 0;j < l;j++) {
            if (b[j].second && a[i].first <= b[j].first && b[j].first <= a[i].second) {
                b[j].second--;
                cnt++;
                break;
            }
        }
    }
    cout << cnt << '\n';
    return 0;
}