POJ·Sunscreen【贪心】

Description–

有C个奶牛去晒太阳 (1 <=C <= 2500)，每只奶牛各自能够忍受的阳光强度有一个minSPF[i]和一个maxSPF[i]值，太大就晒伤了，太小奶牛没感觉。

而刚开始的阳光的强度非常大，奶牛都承受不住，然后奶牛就得涂抹防晒霜，防晒霜的作用是让阳光照在身上的阳光强度固定为某个值。

那么为了不让奶牛烫伤，又不会没有效果，给出了L种防晒霜 (1 <=L <= 2500)。每种的数量cover[i]和固定的阳光强度SPF[i]也给出来了

每个奶牛只能抹一瓶防晒霜，最后问能够享受阳光浴的奶牛有几只。

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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

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Sample Input–

3 2
3 10
2 5
1 5
6 2
4 1

Sample Output–

2

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解题思路–

先排个序~~

• 1.按照最小值递减的顺序把奶牛排序
• 2.按照防晒霜值递减的顺序把防晒霜排序

然后 举个栗子

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代码–

#include<algorithm>
#include<iostream>
#include<cstdio>
using namespace std;
int c,l,ans;
struct cl
{
	int x,y;
}cc[2505],ll[2505];
bool clc(cl xx,cl yy)
{
	return xx.x>yy.x;
}
int main()
{
	scanf("%d%d",&c,&l);
	for (int i=1;i<=c;++i)
	  scanf("%d%d",&cc[i].x,&cc[i].y);
	for (int i=1;i<=l;++i)
	  scanf("%d%d",&ll[i].x,&ll[i].y);
	sort(cc+1,cc+c+1,clc);
	sort(ll+1,ll+l+1,clc);
	for (int i=1;i<=c;++i)
  	  for (int j=1;j<=l;++j)
		if (ll[j].x>=cc[i].x && ll[j].x<=cc[i].y && ll[j].y)//可以抹
		{
		  	ll[j].y--;
		  	ans++;
		  	break;
		}
	printf("%d",ans);
	
	return 0;
}