[AcWing 4480] 倒垃圾

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点击查看代码

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;

const int N = 2e5 + 10;

int n, m;
int x[N], t[N];
vector<int> a, b, c;

int findl(int x)
{
    int l = 0, r = m - 1;
    while (l < r) {
        int mid = l + r + 1 >> 1;
        if (b[mid] <= x)
            l = mid;
        else
            r = mid - 1;
    }
    return l;
}

int findr(int x)
{
    int l = 0, r = m - 1;
    while (l < r) {
        int mid = l + r >> 1;
        if (b[mid] >= x)
            r = mid;
        else
            l = mid + 1;
    }
    return l;
}

void solve()
{
    cin >> n >> m;
    for (int i = 0; i < n + m; i ++)
        cin >> x[i];
    for (int i = 0; i < n + m; i ++)
        cin >> t[i];
    for (int i = 0; i < n + m; i ++) {
        if (t[i] == 1) {
            b.push_back(x[i]);
            c.push_back(0);
        }
        else
            a.push_back(x[i]);
    }
    for (int i = 0; i < n; i ++) {
        int l = findl(a[i]), r = findr(a[i]);
        if (abs(b[l] - a[i]) <= abs(b[r] - a[i]))
            c[l] ++;
        else
            c[r] ++;
    }
    for (int i = 0; i < m; i ++)
        cout << c[i] << ' ';
    cout << endl;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    solve();

    return 0;
}

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1. 用 $a$ 数组来记录所有居民住所的坐标，$b$ 数组来记录所有垃圾桶的坐标
   对于每一个 $a_i$，用两次二分分别找到满足 $b_j \leqslant a_i$ 的最大的 $j$ 和 $b_k \geqslant a_i$ 的最小的 $k$，根据题意条件，如果 $|b_j - a_i| \leqslant |b_k - a_i|$，垃圾应当倒在 $j$，否则，垃圾应当倒在 $k$