Codeforces Round #480 (Div. 2) C. Posterized

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本文介绍了一种海报化滤镜算法的实现方法，旨在帮助学生理解如何将图像颜色范围划分成连续的颜色组，并选择每组的关键颜色来保留图像细节。通过示例详细解释了如何最小化最终颜色数组的方法。

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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.

题目意思大致为：

给你n个数（<=100000个），0<=每个数<=255

问你怎么划分区间，区间为m，要求划出来的区间左端点越小越好。

重点在于并不是把所有的数据都算出，100000很大了

只要得到像素0-255时划分区间的左端点

从头往后找，如果0-255都找到了，那么b[a[i]]!=-1,就不用寻找

#include<bits/stdc++.h>
using namespace std;
const int maxn=100005;
int a[maxn];
int b[maxn];
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        memset(a,0,sizeof(a));
        memset(b,-1,sizeof(b));
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        for(int i=1;i<=n;i++)
        {
            if(b[a[i]]==-1)
            {
                int t=a[i]-m+1;
                for(int j=t;j<=a[i];j++)
                {
                    if(j<0)
                     continue;
                    if(b[j]==-1||b[j]==j)
                    {
                        t=j;
                        break;
                    }
                }
                for(int j=t;j<=a[i];j++)
                    b[j]=t;
            }
        }
        for(int i=1;i<=n;i++)
        {
            if(i==1) printf("%d",b[a[i]]);
            else printf(" %d",b[a[i]]);
        }
        printf("\n");
    }
    return 0;
}