[POJ](3614)Sunscreen ---- 优先级队列+贪心

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

新知：
学习使用优先级队列
参考:
http://www.hankcs.com/program/cpp/poj-3614-sunscreen.html

AC代码：

#include<iostream>#include<cstdio>#include<cstring>#include<queue>#include<algorithm>#include<functional>using namespace std;const int MAX = 2505;typedef struct Cow{    int l;    int h;}node1;node1 cow[MAX];typedef struct SPF{    int p;    int num;}node2;node2 spf[MAX];bool cmp1(node1 a,node1 b){    return a.l<b.l;}bool cmp2(node2 a,node2 b){    return a.p<b.p;}int main(){    priority_queue<int,vector<int>,greater<int> >qi;//值越小的数据优先级越高    ios_base::sync_with_stdio(false);    cin.tie(NULL),cout.tie(NULL);    int C,L;    cin>>C>>L;    for(int i=0;i<C;i++)        cin>>cow[i].l>>cow[i].h;    for(int i=0;i<L;i++)        cin>>spf[i].p>>spf[i].num;    sort(cow,cow+C,cmp1);    sort(spf,spf+L,cmp2);    int now = 0;    int ans = 0;    for(int i=0;i<L;i++)    {        while(now<C && cow[now].l<=spf[i].p)        {            qi.push(cow[now].h);            now++;        }        while(spf[i].num && !qi.empty())        {            int t = qi.top();            qi.pop();            if(t>=spf[i].p)            {                ans++;                spf[i].num--;            }        }    }    cout<<ans<<endl;    return 0;}