POJ 3614 Sunscreen

Sunscreen

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2616		Accepted: 925

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

一道有点难的贪心加优先队列问题

具体思路在代码的注释中有详细说明

#include<iostream>
#include<cstdio>
#include<algorithm>

using namespace std;

typedef struct
{
    int minspf;
    int maxspf;
} COW;

typedef struct
{
    int spf;
    int cover;
} LOTION;

int c,l,sz=0;
COW cow[2550],heap[2550];
LOTION lotion[2550];

bool cmp_cow(COW a,COW b)
{
    return a.minspf<b.minspf;
}

bool cmp_lotion(LOTION a,LOTION b)
{
    return a.spf<b.spf;
}

//对牛维护一个小根堆，键值为所需防晒霜最大能力
void push(COW x)
{
    int i=sz++;
    while(i>0)
    {
        int p=(i-1)/2;
        if(heap[p].maxspf<=x.maxspf)
            break;
        heap[i]=heap[p];
        i=p;
    }
    heap[i]=x;
}

COW pop()
{
    COW ret=heap[0];
    COW x=heap[--sz];
    int i=0;
    while(i*2+1<=sz)
    {
        int a=i*2+1,b=i*2+2;
        if(b<sz&&heap[b].maxspf<heap[a].maxspf)
            a=b;
        if(heap[a].maxspf>=x.maxspf)
            break;
        heap[i]=heap[a];
        i=a;
    }
    heap[i]=x;
    return ret;
}

int main()
{
    while(scanf("%d %d",&c,&l)==2)
    {
        sz=0;
        for(int i=0;i<c;i++)
            scanf("%d %d",&cow[i].minspf,&cow[i].maxspf);
        for(int i=0;i<l;i++)
            scanf("%d %d",&lotion[i].spf,&lotion[i].cover);
        sort(cow,cow+c,cmp_cow);//将所有牛按它们所需防晒霜的最小防晒能力排序
        sort(lotion,lotion+l,cmp_lotion);//将将所有防晒霜按它们的防晒能力排序
        int ans=0,t=0;
        for(int i=0;i<l;i++)
        {
            while(t<c&&lotion[i].spf>=cow[t].minspf)//对于每一种防晒霜，先将所有最小值比它大的牛插入堆中
            {
                push(cow[t]);
                t++;
            }
            while(sz>0&&lotion[i].spf>heap[0].maxspf)//再将所有最大值比它小的牛出堆
                pop();
            while(lotion[i].cover>0&&sz>0)//最后将所有它能保护的牛出堆并将答案+1
            {
                pop();
                ans++;
                lotion[i].cover--;
            }
        }
        printf("%d
",ans);
    }

    return 0;
}