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 ≤ minSPFi ≤ 1,000; minSPFi ≤ maxSPFi ≤ 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 SPFi (1 ≤ SPFi ≤ 1,000). Lotion bottle i can cover coveri 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?
有C个奶牛去晒太阳 (1 <=C <= 2500)，每个奶牛各自能够忍受的阳光强度有一个最小值和一个最大值，太大就晒伤了，太小奶牛没感觉。
而刚开始的阳光的强度非常大，奶牛都承受不住，然后奶牛就得涂抹防晒霜，防晒霜的作用是让阳光照在身上的阳光强度固定为某个值。
那么为了不让奶牛烫伤，又不会没有效果。
给出了L种防晒霜。每种的数量和固定的阳光强度也给出来了
每个奶牛只能抹一瓶防晒霜，最后问能够享受晒太阳的奶牛有几个。
输入输出格式
输入格式：
* Line 1: Two space-separated integers: C and L
* Lines 2..C+1: Line i describes cow i‘s lotion requires with two integers: minSPFi and maxSPFi
* Lines C+2..C+L+1: Line i+C+1 describes a sunscreen lotion bottle i with space-separated integers: SPFi and coveri
输出格式：
A single line with an integer that is the maximum number of cows that can be protected while tanning
输入输出样例
输入样例#1：
复制
3 2 3 10 2 5 1 5 6 2 4 1
输出样例#1： 复制
2
分析
中文翻译有点误导，自己看清楚了，防晒霜是先读防晒指数，再读防晒霜的个数。奶牛按上限排序，而防晒霜直接按防晒值排序就可以了。
我对于这种排序方式的理解是：我们使得防晒值与奶牛上限越接近，意味着后面的防晒霜越难来满足这只奶牛，也就是这瓶防晒霜来满足这只奶牛是更优的。一瓶防晒霜最多只能满足一只奶牛，所以如果能满足，那么就用这瓶防晒霜。这就是这道题一个贪心的想法。
程序
1 #include <bits/stdc++.h>
2 using namespace std;
3 const int MAXC = 2500 + 1;
4 struct node
5 {
6 int Mn, Mx;
7 }cows[MAXC];
8 struct sunscreen
9 {
10 int amt, cov;
11 }ss[MAXC];
12 bool comp1(node x, node y)
13 {
14 return x.Mx < y.Mx;
15 }
16 bool comp2(sunscreen x, sunscreen y)
17 {
18 return x.cov < y.cov;
19 }
20 int main()
21 {
22 int c,l,ans;
23 cin >> c >> l;
24 for (int i = 1; i <= c; i++)
25 cin >> cows[i].Mn >> cows[i].Mx;
26 for (int i = 1; i <= l; i++)
27 cin >> ss[i].cov >> ss[i].amt;
28 sort(cows+1, cows+(c+1),comp1);
29 sort(ss+1, ss+(l+1), comp2);
30 for (int i = 1; i <= c; i++)
31 {
32 for (int j = 1; j <= l; j++)
33 if(ss[j].cov >= cows[i].Mn && ss[j].cov <= cows[i].Mx && ss[j].amt)
34 {
35 ss[j].amt--;
36 ans++;
37 break;
38 }
39 }
40 cout << ans << endl;
41 return 0;
42 }
原文地址：https://www.cnblogs.com/OIerPrime/p/8504387.html