C. Posterized（Div.2）（并查集）

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链接：http://codeforces.com/contest/980/problem/C

Professor Ibrahim has prepared the final homework for his algorithm’s class. He asked his students to implement the Posterization Image Filter.

Their algorithm will be tested on an array of integers, where the $i$ -th integer represents the color of the $i$ -th pixel in the image. The image is in black and white, therefore the color of each pixel will be an integer between 0 and 255 (inclusive).

To implement the filter, students are required to divide the black and white color range [0, 255] into groups of consecutive colors, and select one color in each group to be the group’s key. In order to preserve image details, the size of a group must not be greater than $k$ , and each color should belong to exactly one group.

Finally, the students will replace the color of each pixel in the array with that color’s assigned group key.

To better understand the effect, here is an image of a basking turtle where the Posterization Filter was applied with increasing $k$ to the right.

$\text{[math]}$

To make the process of checking the final answer easier, Professor Ibrahim wants students to divide the groups and assign the keys in a way that produces the lexicographically smallest possible array.

Input

The first line of input contains two integers $n$ and $k$ ( $1 \leq n \leq 10^{5}$ , $1 \leq k \leq 256$ ), the number of pixels in the image, and the maximum size of a group, respectively.

The second line contains $n$ integers $p_{1}, p_{2}, \dots, p_{n}$ ( $0 \leq p_{i} \leq 255$ ), where $p_{i}$ is the color of the $i$ -th pixel.

Output

Print $n$ space-separated integers; the lexicographically smallest possible array that represents the image after applying the Posterization filter.

Examples
inputCopy
4 3
2 14 3 4
outputCopy
0 12 3 3
inputCopy
5 2
0 2 1 255 254
outputCopy
0 1 1 254 254

Note

One possible way to group colors and assign keys for the first sample:

Color $2$ belongs to the group $[0, 2]$ , with group key $0$ .

Color $14$ belongs to the group $[12, 14]$ , with group key $12$ .

Colors $3$ and $4$ belong to group $[3, 5]$ , with group key $3$ .

Other groups won't affect the result so they are not listed here.

大意：在区间[0,255]中，连续的数字分组，每组不多于k个成员，取每组中最小的那个值输出，求总字典序最小的输出

思路：利用并查集，[0,255]的每个值都计数为1

1、如果该值与父节点不相等，则直接输出父节点（因为已经分好组了）

2、如果该值的计数与该值前面元素的父节点计数相加大于m，则直接输出该值（因为这两个联立计数会大于m）

3、如果该值的计数与该值前面元素的父节点计数相加小于等于m，则该值可以与前面元素联立

以下是AC代码：

#include <cstdio>   
#include <iostream> 
#include <algorithm> 
#include <cmath> 
#include <cstdlib> 
#include <cstring> 
#include <vector> 
#include <list> 
#include <map> 
#include <stack> 
#include <queue> 
using namespace std; 
#define ll long long
int f[100005];
int sum[100005];
int n,m;
void set()
{
	for(int i = 0;i <= 260;i++)
	{
		f[i] = i;
		sum[i] = 1;
	}
}
int father(int x)
{
	if(f[x] == x)
		return x;
	else
		return f[x] = father(f[x]);
}
void Union(int x,int y)
{
	int f1 = father(x);
	int f2 = father(y);
	if(f1 != f2)
	{
		f[max(f1,f2)] = min(f1,f2);
		sum[min(f1,f2)]++;
	}
}
int main() 
{
	while(cin >> n >> m)
	{
		set();
		while(n--)
		{
			int x;
			cin >> x;
			while(x - 1 >= 0)
			{
				if(father(x) != x)
					break;
				else if(sum[father(x-1)]+sum[f[x]] > m)
					break;
				else
				{
					sum[father(x-1)]+=sum[f[x]];
					f[x] = father(x-1);
				}
				x--;
			}
			cout << f[x];
			if(n == 0)
				cout << endl;
			else
				cout << " ";
		}
	}
    //cout << "AC" <<endl; 
    return 0; 
}

过了很久，又遇到了这题，找了很久bug，以下是AC代码二：

#include <cstdio>  
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <list>
#include <map>
#include <stack>
#include <queue>
using namespace std;
#define ll long long
int f[270];
int n;
int cent[270];
void init()
{
	for(int i = 0;i <= 256;i++)
	{
		f[i] = i;
		cent[i] = 1;
	}
}
int find(int x)
{
	if(f[x] == x)
		return x;
	return f[x] = find(f[x]);
}
void unio(int x,int y)
{
	int f1 = find(x);
	int f2 = find(y);
	if(f1 != f2)
		f[max(f1,f2)] = min(f1,f2);
}
int main()
{
	int n,k;
	while(scanf("%d%d",&n,&k) != EOF)
	{
		init();
		int a[n];
		int ans[n];
		for(int i = 0;i < n;i++)
			scanf("%d",&a[i]);
		for(int i = 0;i < n;i++)
		{
			for(int j = a[i]-1;j >= 0&& j >= a[i]-k;j--)
				if(cent[find(j)]+cent[find(a[i])] <= k && find(j) != find(a[i]))//当时没有写find(j) != find(a[i])，这样可能就会出现两个相同组再联立一次的情况，导致cent无故加倍（找了很久很久才找到的bug）
				{
					int x1 = find(a[i]);
					int x2 = find(j);
					unio(x1,x2);
					cent[find(a[i])] = cent[x1]+cent[x2];
				}
				else
					break;
			ans[i] = find(a[i]);
		}
		for(int i = 0;i < n-1;i++)
			cout << ans[i] << ' ';
		cout << ans[n-1] <<endl;
	}
	//cout << "AC" <<endl;
	return 0;
}

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