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[],s[];
 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=;i<c;i++)
         scanf("%d%d",&cow[i].a,&cow[i].b);
     for(j=;j<l;j++)
         scanf("%d%d",&s[j].a,&s[j].b);
     sort(cow,cow+c,cmp);
     sort(s,s+l,cmp);
     j=;
     int sum=;
     for(i=;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\n",sum);
 }