Sunscreen

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?

输入

* 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

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

样例输入

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

样例输出

2

分析：把奶牛按最小值排序，防晒霜按固定值排序，从最小的防晒霜枚举，满足条件即放入优先队列。再将最小值取出，更新ans。

#include <iostream>
#include <string>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <deque>
#include <map>
#define range(i,a,b) for(int i=a;i<=b;++i)
#define LL long long
#define rerange(i,a,b) for(int i=a;i>=b;--i)
#define fill(arr,tmp) memset(arr,tmp,sizeof(arr))
using namespace std;
int C,L;
pair<int,int> fuck[2505],you[2505];
priority_queue <int, vector<int>, greater<int> > q;
bool cmp(pair<int,int>a,pair<int,int>b){
    return a.first<b.first;
}
void init() {
    cin >> C >> L;
    range(i, 0, C - 1)cin >> fuck[i].first >> fuck[i].second;
    range(i, 0, L - 1)cin >> you[i].first >> you[i].second;
    sort(fuck,fuck+C,cmp);
    sort(you,you+L,cmp);
}
void solve(){
    int j=0,ans=0;
    range(i,0,L-1){
        while(j<C&&fuck[j].first<=you[i].first){
            q.push(fuck[j].second);
            ++j;
        }
        while(!q.empty()&&you[i].second){
            int tmp=q.top();
            q.pop();
            if(tmp<you[i].first)continue;
            ++ans;
            you[i].second--;
        }
    }
    cout<<ans<<endl;
}
int main() {
    init();
    solve();
    return 0;
}