【线段树求区间第一个不大于val的值】Lpl and Energy-saving Lamps

https://nanti.jisuanke.com/t/30996

线段树维护区间最小值，查询的时候优先向左走，如果左边已经找到了，就不用再往右了。

一个房间装满则把权值标记为INF，模拟一遍，注意考虑一个月内装满多个房间和装满所有房间后不用再购买的情况。

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
const int maxn = 100005;
const int INF = 0x3f3f3f3f;
using namespace std;
int a[maxn], St[maxn << 2], Q[maxn], ans[maxn], remain[maxn];
void PushUp(int rt) {
    St[rt] = min(St[rt << 1], St[rt << 1 | 1]);
}
void Build(int l, int r, int rt) {
    if (l == r) {
        St[rt] = a[l];
        return;
    }
    int m = (l + r) >> 1;
    Build(l, m, rt << 1);
    Build(m + 1, r, rt << 1 | 1);
    PushUp(rt);
}
void Update(int L, int C, int l, int r, int rt) {
    if (l == r) {
        St[rt] = C;
        return;
    }
    int m = (l + r) >> 1;
    if (L <= m) {
        Update(L, C, l, m, rt << 1);
    }
    else {
        Update(L, C, m + 1, r, rt << 1 | 1);
    }
    PushUp(rt);
}
int Query(int val, int L, int R, int l, int r, int rt) {
    if (l == r) {
        if (St[rt] <= val) {
            return l;
        }
        return INF;
    }
    if (L <= l && R >= r) {
        if (St[rt] > val) {
            return INF;
        }
    }
    int m = (l + r) >> 1;
    int ANS = INF;
    if (L <= m) ANS = min(ANS, Query(val, L, R, l, m, rt << 1));
    if (ANS != INF) {
        return ANS;
    }
    if (R > m) ANS = min(ANS, Query(val, L, R, m + 1, r, rt << 1 | 1));
    return ANS;
}
int main() {
    int n, m, q, mxq = 0;
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
    }
    scanf("%d", &q);
    for (int i = 1; i <= q; i++) {
        scanf("%d", &Q[i]);
        mxq = max(mxq, Q[i]);
    }
    Build(1, n, 1);
    for (int i = 1, now = m, fin = 0; i <= mxq; i++, fin >= n ? 0 : now += m) {
        int p = Query(now, 1, n, 1, n, 1);
        ans[i] = ans[i - 1];
        while (p != INF) {
            fin++;
            ans[i]++;
            now -= a[p];
            Update(p, INF, 1, n, 1);
            p = Query(now, 1, n, 1, n, 1);
        }
        remain[i] = now;
    }
    for (int i = 1; i <= q; i++) {
        printf("%d %d\n", ans[Q[i]], remain[Q[i]]);
    }
}

原文地址：https://www.cnblogs.com/stolf/p/9572833.html