POJ 3614 Sunscreen（贪心+优先队列）

Sunscreen

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7131 Accepted: 2518

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 23 102 51 56 24 1

Sample Output

2

Source

USACO 2007 November Gold

题意：有C头牛要进行日光浴，每头牛都有一定的阳光承受范围minSPF_{i到maxSPFi，  现在有L种防晒霜，每种防晒霜可以将日光固定到SPFi，每种防嗮霜的数量为coveri，问最多有多少牛可以沐浴到合适的阳光}

_{思路:要用优先队列来维护牛的最大承受能力，先让最小承受能力符合的进队列，让最大承受能力小的先出队列，其}

_{他的牛可以选择其他的防晒霜}

#include<stdio.h>#include<queue>#include<string.h>#include<algorithm>using namespace std;struct node1{int left,right; } t1[2600]; struct node2 { int score; int num; }t2[2600];bool cmp1(node1 a,node1 b){return a.left<b.left;}bool cmp2(node2 a,node2 b){return a.score<b.score;}int main(){int n,m,i,j,k,l;while(scanf("%d%d",&n,&m)!=EOF){for(i=0;i<n;i++)scanf("%d%d",&t1[i].left,&t1[i].right);for(i=0;i<m;i++)scanf("%d%d",&t2[i].score,&t2[i].num);sort(t1,t1+n,cmp1);sort(t2,t2+m,cmp2);priority_queue<int ,vector<int>,greater<int> >q;//优先队列来维护牛的//最大承受阳光的能力 j=0;int ans=0;for(i=0;i<m;i++){while(j<n&&t1[j].left<=t2[i].score)//然后最小承受能力满足的进队列 {q.push(t1[j].right);j++;}while(!q.empty()&&t2[i].num){int now=q.top();//让最大承受能力最小的出队列 q.pop();if(now>=t2[i].score){ans++;t2[i].num--;}}}printf("%d\n",ans);}return 0;}