HDUoj 4268 Alice and Bob ( STL二分贪心

题目描述

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.

输入

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.

输出

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

样例

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

1
2

题意

给两组 矩形 让我们查找 第一矩形最多有几个比第二组大？

明显的贪心 先将矩形按长x升序排列 然后每次 查找 符合 A[i].x>B[j].x 条件的 B组矩形 扔进multiset里面
multiset里面每次进行二分查找比 A[i].y 大的数 找到了在集合中删去 计数++

AC代码

#include
using namespace std;

#define LL long long
#define CLR(a,b) memset(a,(b),sizeof(a))

const int MAXN = 1e5+111;
multiset<int> mst;
multiset<int>::iterator it;
struct node
{
    int x, y;
    friend bool operator < (node a, node b) {
        return a.x < b.x;
    }
}A[MAXN],B[MAXN];
int main()
{
    ios::sync_with_stdio(false);
    int n, T;
    cin >> T;
    while(T--) {
        mst.clear();
        cin >> n;
        for(int i = 0; i < n; i++) cin >> A[i].x >> A[i].y;
        for(int i = 0; i < n; i++) cin >> B[i].x >> B[i].y;
        sort(A, A+n);
        sort(B, B+n);
        int k = 0;
        for(int i = 0,j = 0; i < n; i++) {
            while(j= B[j].x) mst.insert(B[j++].y);
            if(mst.empty()) continue;
            it = mst.upper_bound(A[i].y);
            if(it != mst.begin()) {
                it--;
                mst.erase(it);
                k++;
            }

        }
        cout << k << endl;
    }


return 0;
}