HDUoj 4268 Alice and Bob ( STL二分贪心

最新推荐文章于 2021-08-13 16:21:43 发布

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最新推荐文章于 2021-08-13 16:21:43 发布

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解决一个游戏中的数学问题，即如何用一组矩形卡片尽可能多地覆盖另一组较小的矩形卡片。采用贪心算法，首先对所有卡片按宽度排序，然后使用二分查找法来确定每个较大卡片可以覆盖的较小卡片。

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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

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<bits/stdc++.h>
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<n && A[i].x >= 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;
}