POJ3614Sunscreen(优先队列+贪心)

Sunscreen

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 8435 Accepted: 2981

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

Source

USACO 2007 November Gold

题目大意:一共有C个牛，每个牛有一个范围，在这个范围内，能够在阳光下晒，否则被烧烤，同时有L个防晒油，每个防晒油有不同的个数和防晒度，也就是说牛抹了防晒油就有防晒油所定的防晒度了，问一个有多少牛可以不被烧烤

解体思路:首先能够想到的是，为了尽可能的让更多的牛能够抹上防晒油，那么我就要尽可能用高于他们忍受范围下线尽可能接近的防晒油摸他们，所以，用优先队列排牛的忍受范围下限从低到高，排列防晒油从低到高，然后从选的那些牛里面找忍受范围上限以下的防晒油给他们

#include<iostream>    #include<cstdio>  #include<stdio.h>  #include<cstring>    #include<cstdio>    #include<climits>    #include<cmath>   #include<vector>  #include <bitset>  #include<algorithm>    #include <queue>  #include<map>  using namespace std;struct cow{int r;int l;}c[2505], cc;struct sunscreen{int sum, ans;}s[2505], ss;priority_queue<int, vector<int>, greater<int> > qua;int n, m, ans;bool cmp(cow x, cow y){return x.l < y.l;}bool cmp1(sunscreen x, sunscreen y){return x.ans < y.ans;}int main(){int i, j;cin >> n >> m;for (i = 1; i <= n; i++){cin >> c[i].l >> c[i].r;}for (i = 1; i <= m; i++){cin >> s[i].ans >> s[i].sum;}sort(c + 1, c + 1 + n,cmp);sort(s + 1, s + 1 + m,cmp1);ans = 0;j = 1;for (i = 1; i <= m; i++){while (j<=n&&c[j].l<=s[i].ans){qua.push(c[j].r);j++;}while(s[i].sum!=0&&!qua.empty()){int k = qua.top(); qua.pop();if (k >= s[i].ans){ans++;s[i].sum--;}}}cout << ans << endl;}