HDU4268 Alice and Bob（贪心+multiset）

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

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一开始用了贪心以后想用alice的最小w从bob最小w开始选，答案当然错了，应该用alice的最小w去选bob尽可能大的w。用了两个for，超时，于是用multiset，让时间复杂度瞬间降低，应该变成O(n)了。

#include<stdio.h>
#include<iostream>
#include<string>
#include<cstring>
#include<algorithm>
#include<queue>
#include<set>

using namespace std;

struct card
{
    int h,w;    
};

int cmp(card a,card b)
{
    if(a.h==b.h)
    {
        return a.w<b.w;
    }
    return a.h<b.h;
}

int main()
{
    int k,m,q,t,p,n;
    int T;
    cin>>T;
    while(T--)
    {
        int ans=0;
        card a[100000],b[100000];
        multiset<int> ms;
        multiset<int>::iterator it;
        
        cin>>n;
        for(int i=0;i<n;++i)
        {
            scanf("%d %d",&a[i].h,&a[i].w);
        }
        for(int i=0;i<n;++i)
        {
            scanf("%d %d",&b[i].h,&b[i].w);
        }
        
        sort(a,a+n,cmp);
        sort(b,b+n,cmp);
        
         int i=0,j=0;
         for(;i<n;++i)
         {
            while(j<n&&(a[i].h>=b[j].h))
            {
                ms.insert(b[j].w);
                ++j;
            }
            if(ms.empty())
            {
                continue;
            }
            
            it=ms.upper_bound(a[i].w);
            if(it==ms.begin())
            {
                continue;
            }else
            {
                --it;
                ++ans;
                ms.erase(it);
            }
            
         }
         cout<<ans<<endl;
         
    }
    return 0;
}