Codeforce 1114 D Flood Fill (DP)

http://codeforces.com/contest/1114/problem/D

D. Flood Fill

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a line of nn colored squares in a row, numbered from 11 to nn from left to right. The ii -th square initially has the color cici .

Let's say, that two squares ii and jj belong to the same connected component if ci=cjci=cj , and ci=ckci=ck for all kk satisfying i<k<ji<k<j . In other words, all squares on the segment from ii to jj should have the same color.

For example, the line [3,3,3][3,3,3] has 11 connected component, while the line [5,2,4,4][5,2,4,4] has 33 connected components.

The game "flood fill" is played on the given line as follows:

• At the start of the game you pick any starting square (this is not counted as a turn).
• Then, in each game turn, change the color of the connected component containing the starting square to any other color.

Find the minimum number of turns needed for the entire line to be changed into a single color.

Input

The first line contains a single integer nn (1≤n≤50001≤n≤5000 ) — the number of squares.

The second line contains integers c1,c2,…,cnc1,c2,…,cn (1≤ci≤50001≤ci≤5000 ) — the initial colors of the squares.

Output

Print a single integer — the minimum number of the turns needed.

Examples

Input

Copy

4
5 2 2 1

Output

Copy

2

Input

Copy

8
4 5 2 2 1 3 5 5

Output

Copy

4

Input

Copy

1
4

Output

Copy

0

Note

In the first example, a possible way to achieve an optimal answer is to pick square with index 22 as the starting square and then play as follows:

• [5,2,2,1][5,2,2,1]
• [5,5,5,1][5,5,5,1]
• [1,1,1,1][1,1,1,1]

In the second example, a possible way to achieve an optimal answer is to pick square with index 55 as the starting square and then perform recoloring into colors 2,3,5,42,3,5,4 in that order.

In the third example, the line already consists of one color only.

区间DP

居然过了。。。

#include<bits/stdc++.h>
using namespace std;
int dp[5005][5005];
int a[5005];
const int inf=0x3f3f3f3f;
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1; i<=n; i++)
        scanf("%d",&a[i]);
    memset(dp,inf,sizeof(dp));
    for(int i=0;i<5005;i++)
        dp[i][i]=0;
    for(int i=n; i>=1; i--)
    {
        for(int j=i; j<=n; j++)
        {
            if(a[i-1]==a[i])
                dp[i-1][j]=min(dp[i][j],dp[i-1][j]);
            else
                dp[i-1][j]=min(dp[i][j]+1,dp[i-1][j]);
            if(a[j+1]==a[j])
                dp[i][j+1]=min(dp[i][j+1],dp[i][j]);
            else
                dp[i][j+1]=min(dp[i][j+1],dp[i][j]+1);
            if(a[i-1]==a[j+1])
                dp[i-1][j+1]=min(dp[i-1][j+1],dp[i][j]+1);
        }
    }
    printf("%d\n",dp[1][n]);
    return 0;
}