hdu4268 Alice and Bob（二维贪心的固定思想 STL的multiset的二分查找函数实现）

Link：http://acm.hdu.edu.cn/showproblem.php?pid=4268

Alice and Bob

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3933    Accepted Submission(s): 1230

Problem Description

Alice and Bob's game never ends. Today, they introduce a new game. In this game, both of them have N different rectangular cards respectively. Alice wants to use his cards to cover Bob's. The card A can cover the card B if the height of A is not smaller than B and the width of A is not smaller than B. As the best programmer, you are asked to compute the maximal number of Bob's cards that Alice can cover.
Please pay attention that each card can be used only once and the cards cannot be rotated.

 

Input

The first line of the input is a number T (T <= 40) which means the number of test cases. 
For each case, the first line is a number N which means the number of cards that Alice and Bob have respectively. Each of the following N (N <= 100,000) lines contains two integers h (h <= 1,000,000,000) and w (w <= 1,000,000,000) which means the height and width of Alice's card, then the following N lines means that of Bob's.

 

Output

For each test case, output an answer using one line which contains just one number.

 

Sample Input

  

   2
2
1 2
3 4
2 3
4 5
3
2 3
5 7
6 8
4 1
2 5
3 4 
  

 

Sample Output

  

   1
2
  

 

Source

2012 ACM/ICPC Asia Regional Changchun Online

 

编程思想： 二维的贪心思想：先找出满足第一维的所有元素，放到集合里面，再找出最接近第二维的元素，然后从集合中删除该元素。

AC code：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
#include<map>
#include<stack>
#include<vector>
#include<set>
#define LL long long
#define MAXN 1000010
using namespace std;
const  int N=20;
struct node{
	int h;
	int w;
	bool operator < (const node &b)const
	{
		if(h!=b.h)
			return h<b.h;
		else
			return w<b.w;
	}
}card1[MAXN],card2[MAXN];
int ans;
multiset<int>sett;
multiset<int>::iterator it;
int main()
{
//	freopen("D:\\in.txt","r",stdin);
	int T,i,j,jj,n;
	scanf("%d",&T);
	while(T--)
	{
		sett.clear();
		scanf("%d",&n);
		for(i=1;i<=n;i++)
		{
			scanf("%d%d",&card1[i].h,&card1[i].w);
		}
		for(i=1;i<=n;i++)
		{
			scanf("%d%d",&card2[i].h,&card2[i].w);
		}
		sort(card1+1,card1+n+1);
		sort(card2+1,card2+n+1);
		j=1;
		ans=0;
		for(i=1;i<=n;i++)
		{
			while(j<=n&&card1[i].h>=card2[j].h)//找出满足第一维的所有元素 
			{
				sett.insert(card2[j].w);//放入集合里面 
				j++;
			}
			if(sett.empty())
				continue;
			it=sett.lower_bound(card1[i].w);//lower_bound(val): 返回容器中第一个值【大于或等于】val的元素的iterator位置 
			if(*it==card1[i].w)//查找的第二维元素满足且最接近（贪心思想：“最接近”在这里的意思是尽可能大的意思） 
			{
				ans++;
				sett.erase(it);
			}
			else//查找的第二维元素不满足，往前找元素 
			{
				if(it==sett.begin()) continue;
				else
				{
					sett.erase(--it);//找第二维满足的且尽可能大的 
					ans++;
				}
			}
		}
		printf("%d\n",ans);
	}
  	return 0;
}