Sunscreen（POJ 3614 优先队列）

Sunscreen

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5898		Accepted: 2068

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

#include <iostream>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
struct node
{
    int a,b;
}cow[2505],s[2505];
bool cmp(node x,node y)
{
    return x.a<=y.a;
}
int main()
{
    int c,l;
    int i,j;
    priority_queue<int,vector<int>,greater<int> > Q;
    freopen("in.txt","r",stdin);
    scanf("%d%d",&c,&l);
    for(i=0;i<c;i++)
        scanf("%d%d",&cow[i].a,&cow[i].b);
    for(j=0;j<l;j++)
        scanf("%d%d",&s[j].a,&s[j].b);
    sort(cow,cow+c,cmp);
    sort(s,s+l,cmp);
    j=0;
    int sum=0;
    for(i=0;i<l;i++)
    {
        while(j<c&&cow[j].a<=s[i].a)
        {
            Q.push(cow[j].b);
            j++;
        }
        while(!Q.empty()&&s[i].b)
        {
            int t=Q.top();
            Q.pop();
            if(s[i].a<=t)
            {
                sum++;
                s[i].b--;
            }
        }
    }
    printf("%d
",sum);
}