Sunscreen

阅读原文时间：2023年07月15日阅读：1

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

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
#include
#include
#include
#include
#include
#include
#include
#include
#include
#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 fuck[],you[];
priority_queue , greater > q;
bool cmp(paira,pairb){
return a.first> C >> L;
range(i, , C - )cin >> fuck[i].first >> fuck[i].second;
range(i, , L - )cin >> you[i].first >> you[i].second;
sort(fuck,fuck+C,cmp);
sort(you,you+L,cmp);
}
void solve(){
int j=,ans=;
range(i,,L-){
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 ;
}

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