POJ3614-Sunscreen-优先队列

原题链接
Sunscreen
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 8079 Accepted: 2861
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
Source

USACO 2007 November Gold

#include <cstdio>#include <iostream>#include <algorithm>#include <vector>#include <queue>using namespace std;const int maxn = 2510;typedef pair<int,int> P;P a[maxn],b[maxn];int main(){    int n,l;    cin >> n >> l;    for(int i=0;i<n;i++) cin >> a[i].first >> a[i].second;    for(int i=0;i<l;i++) cin >> b[i].first >> b[i].second;    sort(a,a+n);    sort(b,b+l);    priority_queue<int, vector<int>, greater<int> > q;    int cnt=0,res=0;    for(int i=0;i<l;i++){        while(cnt < n && a[cnt].first <= b[i].first){            q.push(a[cnt].second);            cnt++;        }        while(!q.empty() && b[i].second){            int t=q.top();            q.pop();            if(t<b[i].first) continue;//消耗后面的没有合适防晒霜的奶牛            res++;            b[i].second--;        }    }    cout << res << endl;    return 0;}