传说的SB DP：
题目
Problem Description
Lee has a string of n pearls. In the beginning, all the pearls have no color. He plans to color the pearls to make it more fascinating. He drew his ideal pattern of the string on a paper and asks for your help.
In each operation, he selects some continuous pearls and all these pearls will be painted to their target colors. When he paints a string which has k different target colors, Lee will cost k2 points.
Now, Lee wants to cost as few as possible to get his ideal string. You should tell him the minimal cost.
In each operation, he selects some continuous pearls and all these pearls will be painted to their target colors. When he paints a string which has k different target colors, Lee will cost k2 points.
Now, Lee wants to cost as few as possible to get his ideal string. You should tell him the minimal cost.
Input
There are multiple test cases. Please process till EOF.
For each test case, the first line contains an integer n(1 ≤ n ≤ 5×104), indicating the number of pearls. The second line contains a1,a2,...,an (1 ≤ ai ≤ 109) indicating the target color of each pearl.
For each test case, the first line contains an integer n(1 ≤ n ≤ 5×104), indicating the number of pearls. The second line contains a1,a2,...,an (1 ≤ ai ≤ 109) indicating the target color of each pearl.
Output
For each test case, output the minimal cost in a line.
Sample Input
3
1 3 3
10
3 4 2 4 4 2 4 3 2 2
Sample Output
2
7
这道题很容易想到方程，但是TLE居多，下面讲讲我的优化！
首先DP方程：DP[J]=MIN(DP[J],DP[I]+SX[I+1,J]^2); I<=J<=N; SX[I,J];表示I-J中有多少个颜色不一样的
我们知道dp[j]最大为J；
所以有假如我们算到DP[I]>=DP[N];那么就可BREAK；
有了这个数据题目就作对了一般，但是遇到全为一样数的数组时，效率不够高
我们需要离散一下。
注意：这份代码在C++ 是TLE ,在G++ 是1S;
对于编译器我无力：
1 #include<stdio.h> 2 #include<string.h> 3 #include<algorithm> 4 #include<math.h> 5 #include<string> 6 #include<map> 7 #include<vector> 8 using namespace std; 9 #define N 55555 10 int b[N],a[N]; 11 int c[N]; 12 int num[N]; 13 int tmp[N]; 14 int dp[N]; 15 vector<int> q; 16 int vis[N]; 17 int main() 18 { 19 int n; 20 while (scanf("%d",&n)!=EOF) 21 { 22 memset(dp,0x3f3f3f,sizeof(dp)); 23 for (int i=1;i<=n;i++) 24 scanf("%d",&a[i]); 25 int t=0; 26 for (int i=1;i<=n;i++) 27 if (b[t]!=a[i]) b[++t]=a[i]; 28 for (int i=1;i<=t;i++) c[i]=b[i]; 29 sort(c+1,c+t+1); 30 int tt=0; 31 for (int i=1;i<=t;i++) 32 if (c[i]!=c[i-1]) a[++tt]=c[i]; 33 for (int i=1;i<=t;i++) c[i]=lower_bound(a+1,a+tt+1,b[i])-a;//离散过程这里用STL处理,前面一段都是把数组离散一下 34 35 36 dp[t]=t;dp[0]=0; 37 memset(vis,0,sizeof(vis)); 38 39 for (int i=0;i<t;i++) 40 { 41 42 for (int j=i+1;j<=t;j++) 43 { 44 if (!vis[c[j]]) vis[c[j]]=1,q.push_back(c[j]); 45 int k=q.size(); 46 if (k*k+dp[i]>=dp[t]) break; 48 dp[j]=min(dp[j],dp[i]+k*k);//没有用long long ,因为超了LONG LONG 就break 49 } 50 51 for (int j=0;j<q.size();j++) 52 vis[q[j]]=0; 53 q.clear(); 54 } 55 printf("%d ",dp[t]); 56 } 57 return 0; 58 }