poj 3614 Sunscreen 贪心

最新推荐文章于 2019-08-27 10:50:57 发布

金编毛

最新推荐文章于 2019-08-27 10:50:57 发布

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分类专栏：贪心文章标签： acm poj

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本文介绍了一种算法，用于解决一群奶牛在海滩上晒太阳时如何合理分配不同SPF值的防晒霜，确保每头奶牛既能避免晒伤又能顺利晒黑的问题。该算法采用优先队列处理，通过对比每头奶牛所需的SPF值范围与现有防晒霜的实际SPF值，以最大化保护晒太阳的奶牛数量。

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Sunscreen

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6189		Accepted: 2147

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

Sample Output

分析：

感觉跟那个汽车加油的题目差不多，用优先队列处理一下就可以啦

ac代码：

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <math.h>
#include <ctype.h>
#include <algorithm>
#include <map>
#include <math.h>
#include <stack>
#include <queue>
#include <set>
#define Max 100010
#define inf 100000000
using namespace std;
pair<int , int> cow[2510],sp[1100];
int main(){
    int C,L;
    while (scanf("%d%d",&C,&L)!=EOF) {
        for (int i=0; i<C; i++) {
            scanf("%d%d",&cow[i].first,&cow[i].second);
        }
        for (int i=0; i<L; i++) {
            scanf("%d%d",&sp[i].first,&sp[i].second);
        }
        sort(cow, cow+C);
        sort(sp, sp+L);
        priority_queue<int,vector<int>,greater<int> > que;
        int j=0,ans= 0;
        for (int i=0; i<L; i++) {
            while (j<C&&cow[j].first<=sp[i].first) {
                que.push(cow[j].second);
                j++;
            }
            while (!que.empty()&&sp[i].second>0) {
                int x = que.top();
                que.pop();
                if (sp[i].first<=x) {
                   // printf("SPF:%d %d\nCow %d\n",sp[i].first,sp[i].second,x);
                    ans++;
                    sp[i].second--;
                }
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}