【poj3614】Sunscreen

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

Source

USACO 2007 November Gold

 

/*
    枚举+优先队列
　　给奶牛的minSPF从小到大排序
    防晒霜从小到大排序 
    从最小的防晒霜枚举
    把minSPF小于防晒霜的牛的maxSPF加到优先队列当中
    优先队列最小值优先，取出最小值。(因为较大的maxSPF有较大的选择空间) 
    更新答案
*/ 

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<queue>
using namespace std;
priority_queue<int> q;
struct cownode
{
    int minSPF,maxSPF;
}cow[2510];//牛 
int sun[1000000];//防晒霜 

int comp(const cownode & a,const cownode & b)
{
    return (a.minSPF<b.minSPF);
}
int main()
{    
    int C,L,SPF,cover,ans=0,j;
    cin>>C>>L;
    for (int i=1;i<=C;i++) cin>>cow[i].minSPF>>cow[i].maxSPF;
    int sunsum=1;
    for (int i=1;i<=L;i++)
    {
        cin>>SPF>>cover;
        for (int j=sunsum;j<=sunsum+cover;j++) sun[j]=SPF;
        sunsum+=cover;
    }
    sunsum--;
    sort(cow+1,cow+C+1,comp);
    sort(sun+1,sun+sunsum+1);
    for (int i=1;i<=sunsum;i++) //枚举防晒霜 
    {
        for (j=1;j<=C;j++)//枚举牛 
            if(cow[j].minSPF<=sun[i])
            {
                q.push(-cow[j].maxSPF);
                cow[j].minSPF=10000000;//给个极大值 再也不会进队列了
            }
        int t=0;
        if (!q.empty()) t=-q.top();//注意是-的q.top(); 
        while (t<sun[i] && !q.empty()) //如果最小的防晒霜也比maxSPF大 这头牛肯定不能晒太阳啦 
        {
             q.pop();//不能在↑取出q.top的时候就删除，当没用的时候再删除！ 
             if (!q.empty()) t=-q.top();
        }
        if (t>=sun[i] && t!=0)
        {
            ans++;
            q.pop();//别忘了删除 
        }
    }
    cout<<ans<<endl;
}