优先队列与贪心 POJ3614

Sunscreen

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 SPFrating 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

代码：

#include<iostream>	//cin和cout函数 
#include<queue>	//优先队列
#include<algorithm>	//sort函数 
#include<cstdio>	//printf和scanf函数 
using namespace std;

struct node { int min,max;	//定义结构体 
	bool operator<(const node &oth)const	//声明优先队列排序方法 
	{
		return min < oth.min;	//每次输出最大元素，针对结构体中的min排序 
	}
} cow[2500];

struct nod1{ int spf,num;}lot[2500];	//结构体 

bool cmp(nod1 x,nod1 y){	//排序声明函数 
	return x.spf>y.spf;	//从大到小 
}

int main(){
	int m,n,i,j,ans=0,min[2500],max[2500],spf[2500],num[2500];
	
	priority_queue < node > dui;	//定义优先队列，node为排序方法 
	
	cin>>m>>n;
	for(i=0;i<m;i++){
		cin>>cow[i].min>>cow[i].max;
		
		node tmp;	//定义结构体，以用来向优先队列中加入一组元素 
		tmp.min=cow[i].min;	//获得数值 
		tmp.max=cow[i].max;
		dui.push(tmp);	//放入 
	}
	for(i=0;i<n;i++)
	cin>>lot[i].spf >>lot[i].num;
	
	sort(lot,lot+n,cmp);	//排序 
	
	for(i=0;i<m;i++){
		node c= dui.top();	//定义结构体，用来取出数据 
		for(j=0;j<n;j++){
			if(lot[j].spf>=c.min&&lot[j].spf<=c.max&&lot[j].num!=0){
				ans++;
				lot[j].num--;
				break;
			}
		}
		dui.pop();	//删除数据 
	}
	printf("%d",ans);
}

    先对所有的奶牛按照最低级从高到低进行优先排序，对
防晒霜按照级别从高到低排序，然后取出第一头奶牛，
用循环在防晒霜中找到最先合适的一瓶，直到循环完所
有奶牛，此题本质还是贪心，只是用到了优先队列的知
识而已，并没有完全发挥其本应有的功能

优先队列与贪心 POJ3614

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