Alice and Bob HDU - 4268

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. 

InputThe 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. 
OutputFor 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

A有n張不同的卡片，B也有n張不同的卡片，其中每個卡片只能用一次（即A的一張卡片只能覆蓋一張B的卡片），問你A最多可以覆蓋B的多少張卡片.

思路：

          對兩個集合按照x進行升序排序，然後對a集合中的每個物品，在x滿足的情況下，將對應的b集合中的物品扔入multiset中，然後在multiset中尋找y值最大的合法值，若找到++ans並將該值erase；

這裏有坑，可以選用upper_bound簡化代碼，在之前應搞清楚upper_bound 與 lower_bound 的區別，推薦一下我的一點博客可以自找一下，還有就是注意迭代器不要越界.

上代碼：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<set>
using namespace std;
int T;
multiset<int> op; 
struct node{
	int x;
	int y;
};
node a[100001];
node b[100001];
bool cmp(node c,node d)
{
	return c.x < d.x;
}
int main()
{
	scanf("%d" ,&T);
	int n;
	while(T--)
	{
	  scanf("%d", &n);
	  for(int i=1; i<=n; i++)
	  {
	    scanf("%d %d",&a[i].x, &a[i].y);		
	  }
	  for(int i=1; i<=n; i++)
	  {
	    scanf("%d %d",&b[i].x,&b[i].y);		
	  }
	  sort(a+1,a+1+n,cmp);
	  sort(b+1,b+1+n,cmp);
	  op.clear();
	  int j = 1;
	  int sum = 0;
	  for(int i=1; i<=n; i++)
	  {
		 for( ; j<=n; j++)
		 {
			if(a[i].x >= b[j].x)
			{
			  op.insert(b[j].y);   	
			}
			else break;
		 }
		 if(op.empty())continue;
		 multiset<int>::iterator it = op.upper_bound(a[i].y);
		 if(it == op.begin()){continue;}
		 if(*--it <= a[i].y){sum++; op.erase(it);}
      }
	printf("%d\n",sum );	
	}
	return 0;
}

水波.