原題鏈接 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 LLines 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 coveriOutput
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;//消耗后面的沒有合適防曬霜的奶牛 res++; b[i].second--; } } cout << res << endl; return 0;}新聞熱點
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