#题解#洛谷P7167 喷泉#ST表#区间最值#

P7167 [eJOI 2020] Fountain (Day1) - 洛谷

分析

1. 由于喷泉确定，比第i个盘子大的第一个盘子nxt i 是确定的。我们由ST表维护nxt i 。

2. 最终落入第几个盘子的答案显然单调，我们进行二分。nxt i j 表示i后面第 j+1个盘子，对 j 二分

代码实现

#include<bits/stdc++.h>
#define int  long long
#define endl '\n'
using namespace std;
const int N = 1e6+10;
int n, q, d[N], c[N], rmax[N][20], f[N][20], g[N][20], log_2[N];
int query_max(int a, int b)
{
	int x = log_2[b - a + 1];
	return max(rmax[a][x], rmax[b - (1 << x) +1][x]);
}
signed main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	cin >> n >> q;
	for (int i = 1; i <= n; i++)
		cin >> d[i] >> c[i];
	for (int i = 2; i <= n; i++)
		log_2[i] = log_2[i >> 1] + 1;
	for (int i = 1; i <= n; i++)
		rmax[i][0] = d[i];
	for (int j = 1; (1 << j) <= n; j++)
		for (int i = 1; i <= n - (1 << j) +1; i++)
			rmax[i][j] = max(rmax[i][j - 1],
			                 rmax[i + (1 << (j - 1))][j - 1]);
	c[n + 1] = 1e9;
	d[n + 1] = 1e9;
	for (int i = 1; i < n; i++)
	{
		int l = i + 1, r = n + 1, mid;
		while (l < r)
		{
			mid = (l + r ) >> 1;
			if (query_max(i + 1, mid) <= d[i])
				l = mid + 1;
			else
				r = mid;
		}
		
		f[i][0] = l;//下一个盘子的编号
		g[i][0] = c[f[i][0]];//下一个盘子的容量
	}
	for (int t = 1; t <= 16; t++)
		for (int i = 1; i <= n; i++)
		{
			f[i][t] = f[f[i][t - 1]][t - 1];//i后第2^t个盘子
			g[i][t] = g[i][t - 1] + g[f[i][t - 1]][t - 1];
		}
	
	while (q--)
	{
		int r, v;
		cin >> r >> v;
		if (v > c[r])
		{
			v -= c[r];
			for (int t = 16; t >= 0; t--)
				if (v > g[r][t])
				{
					v -= g[r][t];
					r = f[r][t];
				}
			r = f[r][0];
		}
		if (r == n + 1) r = 0;
		cout << r << endl;
	}
	return 0;
}