[BZOJ3545] [ONTAK2010]Peaks（线段树合并 + 离散化）

Posted 2020-10-06 蒟蒻zht的博客

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由于困难值小于等于x这个很恶心，可以离线处理，将边权，和询问时的x排序。

每到一个询问的时候，将边权小于等于x的都合并起来再询问。

。。

有重复元素的线段树合并的时间复杂度是nlog^2n

#include <cstdio>
#include <iostream>
#include <algorithm>
#define N 500001

int n, m, q, cnt, tot, size;
int sum[N * 10], ls[N * 10], rs[N * 10], a[N], b[N], f[N], root[N], ans[N], c[N << 1];

struct node
{
	int x, y, z, id;
	node(int x = 0, int y = 0, int z = 0, int id = 0) : x(x), y(y), z(z), id(id) {}
}p[N], ask[N];

inline int read()
{
	int x = 0, f = 1;
	char ch = getchar();
	for(; !isdigit(ch); ch = getchar()) if(ch == ‘-‘) f = -1;
	for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - ‘0‘;
	return x * f;
}

inline bool cmp1(node x, node y)
{
	return x.z < y.z;
}

inline bool cmp2(node x, node y)
{
	return x.y < y.y;
}

inline void merge(int &x, int y)
{
	if(!x || !y)
	{
		x += y;
		return;
	}
	sum[x] += sum[y];
	merge(ls[x], ls[y]);
	merge(rs[x], rs[y]);
}

inline void insert(int &now, int l, int r, int x)
{
	now = ++size;
	if(l == r)
	{
		sum[now] = 1;
		return;
	}
	int mid = (l + r) >> 1;
	if(x <= mid) insert(ls[now], l, mid, x);
	else insert(rs[now], mid + 1, r, x);
	sum[now] = sum[ls[now]] + sum[rs[now]];
}

inline int query(int now, int l, int r, int x)
{
	if(l == r) return l;
	int mid = (l + r) >> 1;
	if(x <= sum[ls[now]])
		return query(ls[now], l, mid, x);
	else
		return query(rs[now], mid + 1, r, x - sum[ls[now]]);
}

inline int find(int x)
{
	return x == f[x] ? x : f[x] = find(f[x]);
}

int main()
{
	int i, j, x, y, z;
	n = read();
	m = read();
	q = read();
	for(i = 1; i <= n; i++) a[i] = b[i] = read();
	std::sort(b + 1, b + n + 1);
	cnt = std::unique(b + 1, b + n + 1) - b - 1;
	for(i = 1; i <= n; i++)
	{
		a[i] = std::lower_bound(b + 1, b + cnt + 1, a[i]) - b;
		f[i] = i;
		insert(root[i], 1, cnt, a[i]);
	}
	for(i = 1; i <= m; i++)
	{
		x = read();
		y = read();
		c[i] = z = read();
		p[i] = node(x, y, z, 0);
	}
	for(i = 1; i <= q; i++)
	{
		x = read();
		c[i + m] = y = read();
		z = read();
		ask[i] = node(x, y, z, i);
	}
	std::sort(c + 1, c + m + q + 1);
	tot = std::unique(c + 1, c + m + q + 1) - c - 1;
	for(i = 1; i <= m; i++)
		p[i].z = std::lower_bound(c + 1, c + tot + 1, p[i].z) - c;
	for(i = 1; i <= q; i++)
		ask[i].y = std::lower_bound(c + 1, c + tot + 1, ask[i].y) - c;
	std::sort(p + 1, p + m + 1, cmp1);
	std::sort(ask + 1, ask + q + 1, cmp2);
	j = 1;
	for(i = 1; i <= q; i++)
	{
		while(j <= m && p[j].z <= ask[i].y)
		{
			x = find(p[j].x);
			y = find(p[j].y);
			if(x ^ y)
			{
				f[y] = x;
				merge(root[x], root[y]);
			}
			j++;
		}
		x = find(ask[i].x);
		if(ask[i].z > sum[root[x]]) ans[ask[i].id] = -1;
		else ans[ask[i].id] = b[query(root[x], 1, cnt, sum[root[x]] - ask[i].z + 1)];
	}
	for(i = 1; i <= q; i++) printf("%d\n", ans[i]);
	return 0;
}