POJ 3614 - Sunscreen

题目链接：http://poj.org/problem?id=3614

Time Limit: 1000MS Memory Limit: 65536K

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

 

把牛按minSPF从小到大排序一下，防晒霜按SPF排序一下。

然后遍历防晒霜lontion[1...L]，对于lotion[i]，把所有满足minSPF小于等于lontion[i].SPF的牛都找出来，入队。

我们将这个队列定义为按cow[i].maxSPF的从小到大排序的优先队列，那么，每次取出的队首，都是当前要求最苛刻（SPF要求范围最小的）那头牛，先满足这些牛，如果当前这 种防晒霜能满足这头牛，就用掉一瓶，否则就放弃这头牛。

反复如此，直到所有防晒霜用完。

#include<cstdio>
#include<algorithm>
#include<queue>
using namespace std;
struct Cows{
    int minSPF,maxSPF;
    bool operator<(const Cows &oth)const
    {
        return maxSPF>oth.maxSPF;
    }
}cow[2505];
struct Lotions{
    int SPF,cover;
}lotion[2505];
bool cmp1(Lotions a,Lotions b) {return a.SPF<b.SPF;}
bool cmp2(Cows a,Cows b) {return a.minSPF<b.minSPF;}
int main()
{
    int C,L;
    priority_queue<Cows> q;
    scanf("%d%d",&C,&L);
    for(int i=1;i<=C;i++) scanf("%d%d",&cow[i].minSPF,&cow[i].maxSPF);
    for(int i=1;i<=L;i++) scanf("%d%d",&lotion[i].SPF,&lotion[i].cover);
    sort(lotion+1,lotion+L+1,cmp1);
    sort(cow+1,cow+C+1,cmp2);
    int now=1,ans=0;
    for(int i=1;i<=L;i++)
    {
        while(now <= C && cow[now].minSPF <= lotion[i].SPF)//由于事先排过序，所以当前这头一旦不满足cow[now].minSPF <= lotion[i].SPF，之后的牛都不满足 
        {
            q.push(cow[now]);
            now++;
        }
        while(!q.empty() && lotion[i].cover>0)//要么是当前这种防晒霜很多，把所有牛都涂了；要么就是当前这种防晒霜用完了
        {
            if(q.top().maxSPF >= lotion[i].SPF)//这种防晒霜可以给这头牛涂
            {
                ans++;//能涂的牛的数量增加1 
                lotion[i].cover--;//当前这种防晒霜用去一瓶
            }
            q.pop();//涂的上就涂，涂不上就放弃
        }
    }
    printf("%d\n",ans);
}