POJ 3614 Sunscreen（优先队列）

Sunscreen
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 8404 Accepted: 2973
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 ≤ minSPFi ≤ 1,000; minSPFi ≤ maxSPFi ≤ 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 SPFi (1 ≤ SPFi ≤ 1,000). Lotion bottle i can cover coveri 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: minSPFi and maxSPFi
• Lines C+2..C+L+1: Line i+C+1 describes a sunscreen lotion bottle i with space-separated integers: SPFi and coveri

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

题意：有C头奶牛日光浴，每头奶牛分别需要minSPF_i和maxSPF_i单位强度之间的阳光。现有L种防晒霜，分别能使阳光强度稳定为SPF_i，其瓶数为cover_i。求最多满足多少头奶牛

题解：将奶牛按照阳光强度的最小值从小到大排序。
将防晒霜也按照能固定的阳光强度从小到大排序
从最小的防晒霜枚举，将所有符合 最小值小于等于该防晒霜的 奶牛的 最大值 放入优先队列之中。
然后优先队列是小值先出

#include <iostream>#include <algorithm>#include <vector>#include <queue>using namespace std;const int maxn = 11110;priority_queue<int,vector<int>,greater<int> >pq;struct node{    int x,y;    friend bool operator < (node a,node b)    {        return a.x < b.x;    }}cow[maxn],bot[maxn];int main(){    int c,l;    cin>>c>>l;    for(int i=0;i<c;i++)        cin>>cow[i].x>>cow[i].y;    for(int i=0;i<l;i++)        cin>>bot[i].x>>bot[i].y;    sort(cow,cow+c);    sort(bot,bot+l);    int cur = 0,ans = 0;// 现在正等待涂防晒霜的奶牛的index    for(int i=0;i<l;i++)    {        while(cur < c&&cow[cur].x <= bot[i].x)        {            pq.push(cow[cur].y);            cur++;        }        while(!pq.empty()&&bot[i].y)        {            int maxSPF = pq.top();            pq.pop();            if(maxSPF >= bot[i].x)// “奶牛上限”比这一瓶的上限大，说明这头奶牛可以被涂上防晒霜            {                ans++;                bot[i].y--;            }// else 这头奶牛不能被涂上，因为bottle是按SPF排过序的，没有比这瓶更小的SPF了        }    }    cout<<ans<<endl;    return 0;}