Alice and Bob（贪心 + multiset）

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本文深入探讨了游戏开发领域的关键技术，包括游戏引擎、编程语言、硬件优化等，并重点阐述了AI音视频处理的应用场景和实现方法，如语义识别、物体检测、语音变声等。通过实例分析，揭示了技术融合带来的创新解决方案。

Alice and Bob

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2571 Accepted Submission(s): 828

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：

首先给出样例个数 T（1 ~ 50），每个样例给出 N （<= 100000）张牌，代表 Alice 和 Bob 各有 N 张牌，每张牌都有长（1 ~ 10 ^ 9）和宽（1 ~ 10 ^ 9），现要用 Alice 的牌去覆盖 Bob 的，必须长和宽都大于等于才能覆盖，问能够覆盖的最大数。

思路：

贪心 + multiset。与昨天的多校贪心题类似，先对牌统一按长由大到小排序，后用 Bob 的牌去匹配 Alice 的牌，选出 Alice 牌中长大于 Bob 长的牌，将宽放入 multiset 中，在从中选出在里面的牌中大于或者等于 Bob 牌的宽的第一个数，运用 upper_bound，之后从 multiset 中删去这张牌即可。用 multiset 的原因是可能宽度会有重复的。寻找的时候统计个数就好了。

lower_bound 返回的是第一个大于或者等于 num的数，若全部都小于则返回 end；

upper_bound 返回的是第一个大于 num 的数，若全部都小于则返回 end。

AC：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>

using namespace std;

const int MAX = 100005;

typedef struct {
        int l, w;
} node;

node a[MAX], b[MAX];

bool cmp (node a, node b) {
        if (a.l != b.l) return a.l > b.l;
        return a.w > b.w;
}

int main() {
        int t;
        scanf("%d", &t);

        while (t--) {
                int n;
                scanf("%d", &n);

                for (int i = 0; i < n; ++i) {
                        scanf("%d%d", &a[i].l, &a[i].w);
                }

                for (int i = 0; i < n; ++i) {
                        scanf("%d%d", &b[i].l, &b[i].w);
                }

                sort(a, a + n, cmp);
                sort(b, b + n, cmp);

                int m = 0, ans = 0;
                multiset<int> s;
                for (int i = 0; i < n; ++i) {
                        while (m != n && a[m].l >= b[i].l) {
                                s.insert(a[m].w);
                                ++m;
                        }

                        multiset<int>::iterator it;
                        it = s.lower_bound(b[i].w);

                        if (it != s.end() && *it >= b[i].w) {
                                ++ans;
                                s.erase(it);
                        }
                }

                printf("%d\n", ans);
        }
}