Sunscreen

Time Limit:1000MS	Memory Limit:65536K
Total Submissions:5713	Accepted:1994

Description

To avoid unsightly burns while tanning, each of theC(1 ≤C≤ 2500) cows must cover her hide with sunscreen when they're at the beach. Cowihas a minimum and maximumSPFrating (1 ≤minSPF_i≤ 1,000;minSPF_i

≤maxSPF_i≤ 1,000) that will work. If theSPFrating is too low, the cow suffers sunburn; if theSPFrating is too high, the cow doesn't tan at all........

The cows have a picnic basket withL(1 ≤L≤ 2500) bottles of sunscreen lotion, each bottleiwith anSPFratingSPF_i(1 ≤SPF_i≤ 1,000). Lotion bottleican covercover_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:CandL
* Lines 2..C+1: Lineidescribes cowi's lotion requires with two integers:minSPF_iandmaxSPF_i

* LinesC+2..C+L+1: Linei+C+1 describes a sunscreen lotion bottleiwith space-separated integers:SPF_iandcover_i

Output

A single line with an integer that is the maximum number of cows that can be protected while tanning

Sample Input

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

Sample Output

2

題意：有C頭奶牛要去沐光浴，太陽光太強烈會晒壞面板，太弱又會沒效果。每頭牛都有一個太陽光適宜的範圍經行沐光浴，分別給出minspf_i和maxspf_i。  有L種防晒霜，每種防晒霜可以把所受陽光固定於一個值spf_i，每種有cover_i瓶。  問最多會有幾頭牛得到合適的光晒強度？ 

題解：woc，USACO的牛生活真是豐富多彩啊(；′⌒`)   貪心策略，在滿足minspf的條件下，儘量將spf的防晒霜塗到maxspf小的奶牛身上，因為maxspf大的奶牛有更多的選擇。這裡就需要一個優先佇列來儲存滿足minspf的奶牛的maxspf的值。     具體解題步驟如下：

1.將奶牛按照minspf升序排列，將防晒霜按照spf升序排列。

2.列舉防晒霜，將minspf<=spf的奶牛的maxspf存到優先佇列中，然後值小的先出佇列，看是否滿足maxspf>=spf，更新記錄值。

程式碼如下：

#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
struct node1
{
	int minspf,maxspf;
}cow[2510];

struct node2
{
	int spf,num;
}lotion[2510];

int cmp1(node1 a,node1 b)
{
	return a.minspf<b.minspf;
}

int cmp2(node2 a,node2 b)
{
	return a.spf<b.spf;
}

int main()
{
	int n,m,i,j;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		for(i=0;i<n;++i)
			scanf("%d%d",&cow[i].minspf,&cow[i].maxspf);
		for(i=0;i<m;++i)
			scanf("%d%d",&lotion[i].spf,&lotion[i].num);
		sort(cow,cow+n,cmp1);
		sort(lotion,lotion+m,cmp2);
		priority_queue<int, vector<int>,greater<int> >q;//維護被選中奶牛的maxspf 
		j=0;
		int ans=0;
		for(i=0;i<m;++i)
		{
			while(j<n&&cow[j].minspf<=lotion[i].spf)
			{
				q.push(cow[j].maxspf);
				j++;
			}
			while(!q.empty()&&lotion[i].num)
			{
				int cnt=q.top();
				q.pop();
				if(cnt>=lotion[i].spf)
				{
					ans++;
					lotion[i].num--;
				}
			}
		}
		printf("%d\n",ans);
	}
	return 0;
}