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

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。

 

?

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#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; }