Alice and Bob HDU4268

Alice and Bob
Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5584    Accepted Submission(s): 1736

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

题意：问A最多覆盖B能覆盖多少。

思路：贪心

#include<set>
#include<map>
#include<stack>
#include<queue>
#include<math.h>
#include<vector>
#include<string>
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
#define maxn 100005
#define maxm 100005
#define mod 1000000007
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
struct node{
    ll x,y;
}a[maxn],b[maxn];
bool cmp(node m,node n){
    if(m.x==n.x)
        return m.y>n.y;
    return m.x>n.x;
}
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        multiset<ll>s;s.clear();
        int n;
        scanf("%d",&n);
        mem(a,0);mem(b,0);
        for(int i=0;i<n;i++)scanf("%lld%lld",&a[i].x,&a[i].y);
        for(int i=0;i<n;i++)scanf("%lld%lld",&b[i].x,&b[i].y);
        sort(a,a+n,cmp);
        sort(b,b+n,cmp);
        int x=0,ans=0;
        for(int i=0;i<n;i++){
            int j=x;
            for(j=x;a[j].x>=b[i].x&&j<n;j++)s.insert(a[j].y);
            x=j;int z=*s.lower_bound(b[i].y);
            if(z>=b[i].y){s.erase(z);}
            else {ans++;}
        }
        printf("%d
",n-ans);
    }
}