Sunscreen

Time Limit: 1000MS Memory Limit: 65536K

Description

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?

Input

  • 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

Output

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

Sample Input

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

Sample Output

2

思路：

本题是一道有限队列的题目，简单的说，乳液的spf要在奶牛的最大的spf和最小的spf之间，首先排序，将奶牛按照最小的spf从小到大排序，乳液按照spf从小到大，然后开始循环，因为之前将奶牛最小的spf从小到大排序了，那么假如奶牛的最小的spf会小于乳液的spf那么，无论后面那种乳液的spf都会符合大于最小spf这个条件，因此就可以将这只奶牛存进优先队列，然后优先队列根据奶牛的 最大的spf从小到大排序，因为小的若是符合，那么大的一定符合，但是小的不符合的情况，那么大的可能可以符合，所以为了是数量最大，那么肯定小的优先，先把难符合条件的先用了，需要注意的是乳液是有数量的，每次用完一瓶，数量要减一（代码虽然长，但是实际真正为核心的就在for循环里面的，并不难理解）。

#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
struct COW {
	int maxn;
	int minn;
};
struct Sunscreen {
	int num;
	int spf;
};
COW cow[2510];
Sunscreen sunscreen[2510];
bool cmp1(COW a, COW b) {
	return a.minn < b.minn;
}
bool cmp2(Sunscreen a, Sunscreen b) {
	return a.spf < b.spf;
}
int main() {
	int C, L;
	cin >> C >> L;
	for (int i = 0; i < C; i++) cin >> cow[i].minn >> cow[i].maxn;
	for (int i = 0; i < L; i++) cin >> sunscreen[i].spf >> sunscreen[i].num;
	sort(cow, cow + C, cmp1);
	sort(sunscreen, sunscreen + L, cmp2);
	int cnt = 0, ans = 0;
	priority_queue<int, vector<int>, greater<int> > q;
	for (int i = 0; i < L; i++) {
		while (cnt < C && cow[cnt].minn <= sunscreen[i].spf) {
			q.push(cow[cnt].maxn);
			cnt++; 
		} 
		while (!q.empty() && sunscreen[i].num) {
			int x = q.top();
			q.pop();
			if (x >= sunscreen[i].spf) {
				sunscreen[i].num--;
				ans++; 
			}
		}
	}
	cout << ans << endl;
	return 0;
}