Poj 3614-Sunscreen

Posted 2020-09-12

Sunscreen

题目链接：http://poj.org/problem?id=3614

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8557		Accepted: 3010

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
大意：略。
方法：把cow按最小值升序排序，bot按spf的大小升序排序。对于每个bot[i].s，如果过cow的最小值都小于它，把这个cow的最大值放入优先队列，然后再从这个优先队列中取出cow的最大值中的最小值，如果
这个最小值还小于bot[i].s,那么直接删除，因为如果它小于bot[i].s，那么它肯定也小于bot[i+1].s;否则，ans+1。
代码：

#include<cstdio>
#include<iostream>
#include<queue>
using namespace std;
const int N=2505;
typedef pair<int,int> P;
P cow[N],bot[N];
bool cmp(P a,P b)
{
    return a.first<b.first;
}
int main()
{
    
    int c,l;
    cin>>c>>l;
    for(int i=0;i<c;i++)
    cin>>cow[i].first>>cow[i].second;
    for(int i=0;i<l;i++)
    cin>>bot[i].first>>bot[i].second;
    sort(cow,cow+c,cmp);
    sort(bot,bot+l,cmp);
    int j=0,ans=0;
    priority_queue<int,vector<int>,greater<int> >q;
    for(int i=0;i<l;i++)
    {
        while(j<c&&cow[j].first<=bot[i].first)
        {
            q.push(cow[j].second); 
            j++;
        }
        while(!q.empty()&&bot[i].second)
        {
            int a=q.top();
            q.pop();
            if(a>=bot[i].first)
            {
                ans++;
                bot[i].second--;
            }
        }
    }
    cout<<ans<<endl;
    return 0;
}