poj 3614 Sunscreen(优先级队列+贪心)

Sunscreen

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 8622 Accepted: 3029

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 23 102 51 56 24 1

Sample Output

2

tips:选取最多的牛涂上防晒霜使其不被晒...题目中要求防晒霜必须在牛的最小值与最大值之间。

贪心选择：每次在防晒霜的可行解中选择牛的最大值最小的，这样才能保证更多的牛涂上防晒霜。

因此将防晒霜以及牛的最小值进行排序，将所有防晒霜大于最小值的牛的最大值入队，

使用优先级队列每次选取其中的最小值，如果防晒霜的值小于最大值，则ans++，同时cover-1；

#include<iostream>#include<utility>#include<cstring>#include<queue>#include<algorithm>using namespace std;typedef pair<int,int>pr;pr cow[2555],scr[2555];int c,l,ans;int main(){cin>>c>>l;for(int i=0;i<c;i++)cin>>cow[i].first>>cow[i].second;for(int i=0;i<l;i++)cin>>scr[i].first>>scr[i].second;sort(cow,cow+c);sort(scr,scr+l);priority_queue<int,vector<int>,greater<int> >q;int j=0;for(int i=0;i<l;i++)//枚举防晒霜 {while(j++<c&&scr[i].first>=cow[j].first)q.push(cow[j].second);while(!q.empty()&&scr[i].second){int x=q.top();q.pop();if(scr[i].first<=x){ans++;scr[i].second--;}}}cout<<ans<<endl;return 0;}