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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;
}
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原文地址：http://www.cnblogs.com/Asimple/p/5487286.html
