[ICPC2020南京L] Let's Play Curling

[ICPC2020南京L] Let's Play Curling - 双指针

Description

给定长度为n的整数数组a，和长度为m的整数数组b，表示在一维数轴上有n个红点 (a[i]),和m个蓝点 (b[i])，给定一个实数c，对于每一个i，如果对于所有的j，都满足 (|c-a[i]|<|c-b[j]|)，那么红队分数+1，现在你可以自定义c的值，计算红队可能获得的最大分数.

Solution

其实就是计算两个蓝点严格夹的区间中最多有多少个红点，双指针即可

#include <bits/stdc++.h>
using namespace std;

#define int long long
const int N = 1e6 + 5;

int n, m, a[N], b[N];

void solve()
{
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    for (int i = 1; i <= m; i++)
        cin >> b[i];
    sort(a + 1, a + n + 1);
    sort(b + 1, b + m + 1);
    a[0] = b[0] = -9e9;
    a[n + 1] = b[m + 1] = 9e9;
    int ans = 0;
    int pos = 1;
    for (int i = 1; i <= m + 1; i++)
    {
        // 可用区间 (b[i-1], b[i])
        int cnt = 0;
        while (a[pos] < b[i] && pos <= n)
        {
            cnt += a[pos] > b[i - 1];
            ++pos;
        }
        ans = max(ans, cnt);
    }
    if (ans)
        cout << ans << endl;
    else
        cout << "Impossible" << endl;
}

signed main()
{
    ios::sync_with_stdio(false);
    int t;
    cin >> t;
    while (t--)
        solve();
}