【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]]);
	}
}
posted @ 2018-09-02 10:06  Stolf  阅读(163)  评论(0编辑  收藏  举报