CodeTON Round 6 (Div. 1 + Div. 2, Rated, Prizes!)C - Colorful Table

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#算法

文章描述了如何通过编程解决一个关于二维数组中寻找最小包含特定颜色区域的矩形，并计算其周长的问题。

C. Colorful Table

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given two integers n� and k�. You are also given an array of integers a1,a2,…,an�1,�2,…,�� of size n�. It is known that for all 1≤i≤n1≤�≤�, 1≤ai≤k1≤��≤�.

Define a two-dimensional array b� of size n×n�×� as follows: bi,j=min(ai,aj)��,�=min(��,��). Represent array b� as a square, where the upper left cell is b1,1�1,1, rows are numbered from top to bottom from 11 to n�, and columns are numbered from left to right from 11 to n�. Let the color of a cell be the number written in it (for a cell with coordinates (i,j)(�,�), this is bi,j��,�).

For each color from 11 to k�, find the smallest rectangle in the array b� containing all cells of this color. Output the sum of width and height of this rectangle.

Input

The first line contains a single integer t� (1≤t≤1041≤�≤104) — the number of test cases. Then follows the description of the test cases.

The first line of each test case contains two integers n� and k� (1≤n,k≤1051≤�,�≤105) — the size of array a� and the number of colors.

The second line of each test case contains n� integers a1,a2,…,an�1,�2,…,�� (1≤ai≤k1≤��≤�) — the array a�.

It is guaranteed that the sum of the values of n� and k� over all test cases does not exceed 105105.

Output

For each test case, output k� numbers: the sums of width and height of the smallest rectangle containing all cells of a color, for each color from 11 to k�.

Example

input

Copy

2 1

1 1

2 2

1 2

3 5

3 2 4

4 2

1 2 1 2

5 3

1 2 3 2 1

output

Copy

Note

In the first test case, the entire array b� consists of color 11, so the smallest rectangle for color 11 has a size of 2×22×2, and the sum of its sides is 44.

In the second test case, the array b� looks like this:

1	1
1	2

One of the corner cells has color 22, and the other three cells have color 11. Therefore, the smallest rectangle for color 11 has a size of 2×22×2, and for color 22 it is 1×11×1.

In the last test case, the array b� looks like this:

1	1	1	1	1
1	2	2	2	1
1	2	3	2	1
1	2	2	2	1
1	1	1	1	1

#include <bits/stdc++.h>

using namespace std;
using ll=long long;
ll LEFT_l[1000010], RIGHT_l[1000010],LEFT_r[1000010], RIGHT_r[1000010];
void findRectangles(ll n, ll k, vector<ll> &a)
{
    for(int i=0; i<max(k,n)+100; i++)
    {
        LEFT_l[i]=1000000;
//        MIN[i]=1000000;
        RIGHT_r[i]=-1;
//        MAX[i]=-1;
    }
    ll MAX=-1,MIN=1000000;
    for (ll i = 0; i < n; i++)
    {
        ll x=a[i];
        LEFT_l[x] = min(LEFT_l[x], i);
        RIGHT_r[x] = max(RIGHT_r[x], i);
    }
    vector<ll>ans;
    ans.clear();
    for(ll i=k; i>0; i--)
    {
        if(LEFT_l[i]==1000000&&RIGHT_r[i]==-1)
        {
            ans.push_back(0);
        }
        else
        {

            MAX=max(MAX,RIGHT_r[i]);
            MIN=min(MIN,LEFT_l[i]);
            ans.push_back(2*(MAX-MIN+1));
        }
    }

    for(int i=0; i<max(n,k)+100; i++)
    {
        LEFT_l[i]=1000000;
//        MIN[i]=1000000;
        RIGHT_r[i]=-1;
//        MAX[i]=-1;
    }

    for (ll i = k-1; i >=0; i--)
    {
        cout<<ans[i]<<' ';
    }
    cout << endl;
}

int main()
{
    ios::sync_with_stdio(false);
    ll t;
    cin >> t;
    while (t--)
    {
        ll n, k;
        cin >> n >> k;

        vector<ll> a(n);
        for (ll i = 0; i < n; i++)
        {
            cin >> a[i];
        }

        findRectangles(n, k, a);
    }

    return 0;
}

清空数组没有清空完全

for(int i=0; i<max(k,n)+100; i++)