Codeforces 911G - Mass Change Queries

G. Mass Change Queries

time limit per test 3 seconds

memory limit per test 512 megabytes

input standard input

output standard output

You are given an array a consisting of n integers. You have to process q queries to this array; each query is given as four numbers l, r, x and y, denoting that for every i such that l ≤ i ≤ r and ai = x you have to set ai equal to y.

Print the array after all queries are processed.

Input

The first line contains one integer n (1 ≤ n ≤ 200000) — the size of array a.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 100) — the elements of array a.

The third line contains one integer q (1 ≤ q ≤ 200000) — the number of queries you have to process.

Then q lines follow. i-th line contains four integers l, r, x and y denoting i-th query (1 ≤ l ≤ r ≤ n, 1 ≤ x, y ≤ 100).

Output

Print n integers — elements of array a after all changes are made.

Example

input

5
1 2 3 4 5
3
3 5 3 5
1 5 5 1
1 5 1 5

output

5 2 5 4 5 

链接：Codeforces 911G - Mass Change Queries

大意：给你n个数，处理q次询问，每次询问要求将l-r之间值为x的数修改为y，最后输出q次询问之后的数列。

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思路：可以利用线段树减少时间复杂度，对于每个节点，储存该节点对应区间内每个数对应的值。

代码：

#include<iostream>

using namespace std;

#define SIZE 200000

inline int lson(int ord) { return ord << 1; }
inline int rson(int ord) { return ord << 1 | 1; }

struct stn 
{
	int left, right;
	int change_to[101];
};

int value[SIZE];
stn nodes[SIZE << 2];

void build(int ord, int l, int r)
{
	nodes[ord].left = l, nodes[ord].right = r;
	for (int i = 0; i < 101; i++)
		nodes[ord].change_to[i] = i;
	if (l == r)
		return;
	int mid = (l + r) >> 1;
	build(lson(ord), l, mid);
	build(rson(ord), mid + 1, r);
}

inline void push_down(int ord)
{
	for (int i = 1; i <= 100; i++)
	{
		nodes[lson(ord)].change_to[i] = nodes[ord].change_to[nodes[lson(ord)].change_to[i]];
		nodes[rson(ord)].change_to[i] = nodes[ord].change_to[nodes[rson(ord)].change_to[i]];
	}
	for (int i = 1; i <= 100; i++)
		nodes[ord].change_to[i] = i;
}

void change_between(int ord, int l, int r, int& x, int& y)
{
	if (l == nodes[ord].left && nodes[ord].right == r)
	{
		for (int i = 1; i <= 100; i++)
			if (nodes[ord].change_to[i] == x)
				nodes[ord].change_to[i] = y;
		return;
	}
	int mid = (nodes[ord].left + nodes[ord].right) >> 1;
	push_down(ord);
	if (r <= mid)
		change_between(lson(ord), l, r, x, y);
	else if (mid < l)
		change_between(rson(ord), l, r, x, y);
	else
	{
		change_between(lson(ord), l, mid, x, y);
		change_between(rson(ord), mid + 1, r, x, y);
	}
}

int get_val(int ord, int pos)
{
	if (nodes[ord].left == pos && pos == nodes[ord].right)
		return nodes[ord].change_to[value[nodes[ord].left]];
	if (pos <= (nodes[ord].left + nodes[ord].right) >> 1)
		return nodes[ord].change_to[get_val(lson(ord), pos)];
	else
		return nodes[ord].change_to[get_val(rson(ord), pos)];
}

int main()
{
	int n, q, l, r, x, y;
	while (scanf("%d", &n) != EOF)
	{
		for (int i = 1; i <= n; i++)
			scanf("%d", &value[i]);
		memset(nodes, 0, sizeof(nodes));
		build(1, 1, n);
		scanf("%d", &q);
		for (int i = 0; i < q; i++)
		{
			scanf("%d %d %d %d", &l, &r, &x, &y);
			change_between(1, l, r, x, y);
		}
		for (int i = 1; i <= n; i++)
			printf("%d%c", get_val(1, i), i == n ? '\n' : ' ');
	}
	return 0;
}