NC216012 Let'sPlayCurling

题目链接

题目

题目描述

Curling is a sport in which players slide stones on a sheet of ice toward a target area. The team with the nearest stone to the center of the target area wins the game.

Two teams, Red and Blue, are competing on the number axis. After the game there are \((n+m)\) stones remaining on the axis, \(n\) of them for the Red team and the other \(m\) of them for the Blue. The iii-th stone of the Red team is positioned at \(a_i\) and the \(i\)-th stone of the Blue team is positioned at \(b_i\).

Let \(c\) be the position of the center of the target area. From the description above we know that if there exists some \(i\) such that \(1 \le i \le n\) and for all \(1 \le j \le m\) we have\(|c - a_i| < |c - b_j|\) then Red wins the game. What's more, Red is declared to win \(p\) points if the number of \(i\) satisfying the constraint is exactly \(p\).

Given the positions of the stones for team Red and Blue, your task is to determine the position \(c\) of the center of the target area so that Red wins the game and scores as much as possible. Note that \(c\) can be any real number, not necessarily an integer.

输入描述

There are multiple test cases. The first line of the input contains an integer \(T\) indicating the number of test cases. For each test case:

The first line contains two integers nnn and mmm (\(1 \le n, m \le 10^5\)) indicating the number of stones for Red and the number of stones for Blue.

The second line contains nnn integers \(a_1, a_2, \cdots, a_n\) (\(1 \le a_i \le 10^9\)) indicating the positions of the stones for Red.

The third line contains mmm integers \(b_1, b_2, \cdots, b_m\) (\(1 \le b_i \le 10^9\)) indicating the positions of the stones for Blue.

It's guaranteed that neither the sum of nnn nor the sum of \(m\) will exceed \(5 \times 10^5\) .

输出描述

For each test case output one line. If there exists some \(c\) so that Red wins and scores as much as possible, output one integer indicating the maximum possible score of Red (NOT \(c\)). Otherwise output "Impossible" (without quotes) instead.

示例1

输入

3
2 2
2 3
1 4
6 5
2 5 3 7 1 7
3 4 3 1 10
1 1
7
7

输出

2
3
Impossible

备注

For the first sample test case we can assign \(c = 2.5\) so that the stones at position 2 and 3 for Red will score.

For the second sample test case we can assign \(c = 7\) so that the stones at position 5 and 7 for Red will score.

题解

知识点：STL，模拟。

选择一个点作为目标，如果存在红色石头比所有蓝色石头都严格接近目标，则红方胜利。进一步，满足比所有蓝色石头都严格接近目标的红色石头的数量是红方最后得分。

显然，红色石头会连成一段区间，中间不能有蓝色石头（包括端点），目标可以选在他们中间，使得这段红色石头都包括其中，并不含蓝色石头，最终得分就是连续红色石头的个数。地图上有很多蓝色石头分割了许多红色石头的区间，所以首先用一个 \(map\) 将坐标映射到红色石头数量，再将蓝色石头坐标的红色石头数量设为 \(0\) ，这样就有了所有石头的坐标，不为 \(0\) 的就是红色石头数量。遍历一遍，计算每个区间的红色石头个数，取其中最大值即可。

时间复杂度 \(O((n+m) \log (n+m))\)

空间复杂度 \(O(n+m)\)

代码

#include <bits/stdc++.h>
#define ll long long

using namespace std;

bool solve() {
    int n, m;
    cin >> n >> m;
    map<int, int> mp;
    for (int i = 0, tmp;i < n;i++) {
        cin >> tmp;
        mp[tmp]++;
    }
    for (int i = 0, tmp;i < m;i++) {
        cin >> tmp;
        mp[tmp] = 0;
    }
    int ans = 0, sum = 0;
    for (auto [i, j] : mp) {
        if (j) sum += j;
        else sum = 0;
        ans = max(ans, sum);
    }
    if (ans) cout << ans << '\n';
    else return false;
    return true;
}

int main() {
    std::ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int t = 1;
    cin >> t;
    while (t--) {
        if (!solve()) cout << "Impossible" << '\n';
    }
    return 0;
}