You are given a line of ci.
Let's say, that two squares j should have the same color.
For example, the line 3 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 1≤n≤5000) — the number of squares.
The second line contains integers 1≤ci≤5000) — the initial colors of the squares.
Output
Print a single integer — the minimum number of the turns needed.
Examples
Input
4 5 2 2 1
Output
2
Input
8 4 5 2 2 1 3 5 5
Output
4
Input
1 4
Output
0
能用区间dp一个很重要的原因是只能通过一个起点更新
代码：
#include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #include<queue> #include<stack> #include<set> #include<map> #include<vector> #include<cmath> const int maxn=5e3+5; typedef long long ll; using namespace std; vector<int>vec; int dp[5005][5005]; int main() { int n; cin>>n; int x; for(int t=1;t<=n;t++) { scanf("%d",&x); vec.push_back(x); } vec.erase(unique(vec.begin(),vec.end()),vec.end()); int nn=vec.size(); memset(dp,0x3f3f3f3f,sizeof(dp)); for(int t=0;t<nn;t++) { dp[t][t]=0; } for(int t=nn-1;t>=0;t--) { for(int len=1;t+len<nn;len++) { if(vec[t]==vec[t+len]) { if(len==1) { dp[t][t+len]=0; } else { dp[t][t+len]=dp[t+1][t+len-1]+1; } } else { dp[t][t+len]=min(dp[t][t+len-1],dp[t+1][t+len])+1; } } } cout<<dp[0][nn-1]<<endl; //system("pause"); return 0; }