ACM题目———

ACM题目————Sunscreen

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

Sample Output

题意

有C个奶牛去晒太阳 (1 <=C <= 2500)，每个奶牛各自能够忍受的阳光强度有一个最小值和一个最大值，太大就晒伤了，太小奶牛没感觉。

而刚开始的阳光的强度非常大，奶牛都承受不住，然后奶牛就得涂抹防晒霜，防晒霜的作用是让阳光照在身上的阳光强度固定为某个值。

那么为了不让奶牛烫伤，又不会没有效果。

给出了L种防晒霜。每种的数量和固定的阳光强度也给出来了

每个奶牛只能抹一瓶防晒霜，最后问能够享受晒太阳的奶牛有几个。

可以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

所以就可以将这些最大值中的最小的取出来。更新答案。

#include <iostream>
#include <algorithm>
#include <queue>

using namespace std;
typedef pair<int,int> p;
priority_queue <int, vector<int>, greater<int> > q;
//定义优先队列，小的先出队
p n[2505], b[2505];

bool cmp(p x, p y)//按pair类的第一个值排序
{
    return x.first<y.first;
}

int main()
{
    int C, L, sum;
    cin >> C >> L ;
    for(int i=0; i<C; i++)
        cin >> n[i].first >> n[i].second ;
    for(int i=0; i<L; i++)
        cin >> b[i].first >> b[i].second ;
    sort(n,n+C,cmp);
    sort(b,b+L,cmp);
    int j = 0 ;
    sum = 0 ;
    for(int i=0; i<L; i++)
    {
        while( j<C && n[j].first<=b[i].first)
        {
            q.push(n[j].second);
            j++ ;
        }
        while(!q.empty() && b[i].second)
        {
            int temp = q.top() ;
            q.pop();
            if(temp < b[i].first) continue ;
            sum ++ ;
            b[i].second -- ;
        }
    }
    cout << sum << endl ;

    return 0;
}

低调做人，高调做事。