POJ 3614 Sunscreen 优先队列

Sunscreen

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 2952 Accepted: 1031

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 ≤ 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 ≤ SPF_i ≤ 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?

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: minSPF_i and maxSPF_i 
* Lines C+2..C+L+1: Line i+C+1 describes a sunscreen lotion bottle i with space-separated integers: SPF_i and cover_i

Output

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

Sample Input

3 23 102 51 56 24 1

Sample Output

2

Source

USACO 2007 November Gold

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每次将“最合适的奶牛”涂上即可

#define  _CRT_SECURE_NO_WARNINGS#include <iostream>#include <cstdio>#include <queue>#include <functional>using namespace std;typedef pair<int, int> P;priority_queue<int, vector<int>, greater<int>> que;P cow[2600], lot[2600];int ans = 0;int main(){int C, L, j = 0;scanf("%d%d", &C, &L);for(int i = 0; i < C; i++)scanf("%d%d", &cow[i].first, &cow[i].second);for(int i = 0; i < L; i++)scanf("%d%d", &lot[i].first, &lot[i].second);sort(cow, cow+C);sort(lot, lot+L);for(int i = 0; i < L; i++){for(; j < C && cow[j].first <= lot[i].first; j++)que.push(cow[j].second);while(lot[i].second && que.size()){int maxspf = que.top();que.pop();if(maxspf >= lot[i].first){ans++;lot[i].second--;}}}printf("%d\n", ans);}