D. Flood Fill

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

题意：连续相同颜色的格子为连通块，选一个起点，每次可以改变相邻的块的颜色，问几次可以使得全部颜色一样；

 思路：1.区间dp，dp[l][r][0]表示l-r颜色为Cl，dp[l][r][1]表示l-r颜色为Cr；

             2.求出最长回文子序列，以此为起点最优；

#include<algorithm>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstring>
#include<sstream>
#include<cstdio>
#include<ctime>
#include<map>
#include<stack>
#include<string>
using namespace std;

#define sfi(i) scanf("%d",&i)
#define pri(i) printf("%d\n",i)
#define sff(i) scanf("%lf",&i)
#define ll long long
#define mem(x,y) memset(x,y,sizeof(x))
#define INF 0x3f3f3f3f
#define eps 1e-6
#define PI acos(-1)
#define lowbit(x) ((x)&(-x))
#define zero(x) (((x)>0?(x):-(x))<eps)
#define fl() printf("flag\n")
ll gcd(ll a,ll b){while(b^=a^=b^=a%=b);return a;}
const int maxn=5e3+9;
const int mod=1e9+7;

template <class T>
inline void sc(T &ret)
{
    char c;
    ret = 0;
    while ((c = getchar()) < '0' || c > '9');
    while (c >= '0' && c <= '9')
    {
        ret = ret * 10 + (c - '0'), c = getchar();
    }
}

int dp[maxn][maxn][2];
int a[maxn],b[maxn];
int cnt;


int main()
{
    int n;
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
    }
    for(int i=1;i<=n;i++)
    {
        if(a[i]!=a[i-1]) b[++cnt]=a[i];
    }
    n=cnt;
    mem(dp,INF);
    for(int i=1;i<=n;i++) dp[i][i][1]=0,dp[i][i][0]=0;
    for(int r=1;r<=n;r++)
    {
        for(int l=r;l>=1;l--)
        {
            dp[l-1][r][0]=min(dp[l-1][r][0],dp[l][r][0]+(b[l-1]!=b[l]));
            dp[l-1][r][0]=min(dp[l-1][r][0],dp[l][r][1]+(b[l-1]!=b[r]));

            dp[l][r+1][1]=min(dp[l][r+1][1],dp[l][r][0]+(b[l]!=b[r+1]));
            dp[l][r+1][1]=min(dp[l][r+1][1],dp[l][r][1]+(b[r]!=b[r+1]));
        }
    }
    cout<<min(dp[1][n][0],dp[1][n][1])<<endl;
    return 0;
}

#include<algorithm>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstring>
#include<sstream>
#include<cstdio>
#include<ctime>
#include<map>
#include<stack>
#include<string>
using namespace std;

#define sfi(i) scanf("%d",&i)
#define pri(i) printf("%d\n",i)
#define sff(i) scanf("%lf",&i)
#define ll long long
#define mem(x,y) memset(x,y,sizeof(x))
#define INF 0x3f3f3f3f
#define eps 1e-6
#define PI acos(-1)
#define lowbit(x) ((x)&(-x))
#define zero(x) (((x)>0?(x):-(x))<eps)
#define fl() printf("flag\n")
ll gcd(ll a,ll b){while(b^=a^=b^=a%=b);return a;}
const int maxn=5e3+9;
const int mod=1e9+7;

template <class T>
inline void sc(T &ret)
{
    char c;
    ret = 0;
    while ((c = getchar()) < '0' || c > '9');
    while (c >= '0' && c <= '9')
    {
        ret = ret * 10 + (c - '0'), c = getchar();
    }
}

int dp[maxn][maxn];
int a[maxn],b[maxn];
int cnt;

void LPS(int len)
{
    for(int i=len;i>0;i--)
    {
        dp[i][i]=1;
        for(int j=i+1;j<=len;j++)
        {
            if(b[i]==b[j])
                dp[i][j]=dp[i+1][j-1]+2;
            else
                dp[i][j]=max(dp[i+1][j],dp[i][j-1]);
        }
    }
}

int main()
{
    int n;
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
    }
    for(int i=1;i<=n;i++)
    {
        if(a[i]!=a[i-1]) b[++cnt]=a[i];
    }
    n=cnt;
    LPS(cnt);
    cout<<n-(dp[1][cnt]+1)/2<<endl;
    return 0;
}