poj 3614 Sunscreen 贪心

Sunscreen

Time Limit: 1000MS Memory Limit: 65536KTotal 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

3 23 102 51 56 24 1

Sample Output

2

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 100000000using 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;}