poj 3614 Sunscreen

                                                                                                    Sunscreen

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7450		Accepted: 2627

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 2
3 10
2 5
1 5
6 2
4 1

Sample Output

2

题意：有C头牛晒太阳，每头牛都有相应的SPF范围，在范围内可以让牛享受太阳，有L瓶防晒霜，每瓶防晒霜都有各自的SPF值，SPF值在某头牛的SPF范围内的防晒霜如果涂在该牛上，可以让该牛享受太阳，问有多少头牛可以享受太阳。
思路：贪心，对于每一种防晒霜，在符合条件的范围内(防晒霜的SPF值大于牛的min_SPF)，每次都筛选出一群牛，将这群牛的max_SPF值压入最小堆，之后尽量用SPF值较小的防晒霜涂于max_SPF较小的牛上。

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<queue>
#include<algorithm>
#include<functional>//greater
using namespace std;
const int N_MAX = ;
struct cows {
    int low;
    int high;
    bool operator <(const cows&b)const{
        return low<b.low||(low==b.low&&high<b.high);
    }
};
struct  lotion {
    int spf;
    int num;
    bool operator <(const lotion&b)const {
        return spf <b.spf ;
    }
};
priority_queue<int,vector<int>,greater<int>>que;//小元素在上
cows cow[N_MAX];
lotion lo[N_MAX];
int main() {
    int C, L;
    scanf("%d%d",&C,&L);
        for (int i = ;i < C;i++)
            scanf("%d%d", &cow[i].low, &cow[i].high);
        for (int i = ;i < L;i++)
            scanf("%d%d", &lo[i].spf, &lo[i].num);
        sort(cow, cow + C);//按牛的spf最小值从小到大排序
        sort(lo, lo + L);//按护肤品的spf值从小到大排
        int result=,cur=;
        for (int i = ;i < L;i++) {//对于每一种护肤品进行考虑
            while (cur<C&&lo[i].spf>=cow[cur].low) {//挑选出满足条件的牛的high值进堆
                que.push(cow[cur].high);//进堆后小的high值在上面
                cur++;
            }
            while (que.size() && lo[i].num) {//如果这种护肤品对所有挑选出来的牛都不满足条件，那么只能弃用，因为后面的牛的min值都比护肤品的spf大
                int k = que.top();que.pop();
                if (k >= lo[i].spf) {//high值小于护肤品的spf值的牛只能不涂，后面的护肤品spf值越来越大更不满足
                    result++;
                    lo[i].num--;
                }
            }
        }
        printf("%d\n",result);
        return ;

}