HDOJ 5009 Paint Pearls, hdoj5009

HDOJ 5009 Paint Pearls, hdoj5009

DP + optimization, because the cost is n ^ 2, so when num × num is greater than DP [I], you can jump out ....

Paint Pearls Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission (s): 1245 Accepted Submission (s): 395


Problem DescriptionLee 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 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 shoshould tell him the minimal cost.
InputThere 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.
OutputFor each test case, output the minimal cost in a line.
Sample Input

31 3 3103 4 2 4 4 2 4 3 2 2

 
Sample Output

27

 
Source2014 ACM/ICPC Asia Regional Xi 'an Online

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <map>using namespace std;int n,a[50050],b[50050],last[50050];int dp[50050];map<int,int> mp;int fa[50050];int find(int x){if(x==fa[x]) return x;return fa[x]=find(fa[x]);}void bing(int x,int y){int X=find(x),Y=find(y);if(X==Y) return ;fa[Y]=X;}int main(){while(scanf("%d",&n)!=EOF){for(int i=1;i<=n;i++){scanf("%d",a+i);b[i]=a[i];}sort(b+1,b+n+1);int cnt=unique(b,b+n+1)-b;for(int i=1;i<cnt;i++){mp[b[i]]=i;}for(int i=0;i<=n;i++){a[i]=mp[a[i]];fa[i]=i;last[i]=-1;}for(int i=1;i<=n;i++){if(last[a[i]]!=-1){bing(last[a[i]]-1,last[a[i]]);}last[a[i]]=i;dp[i]=i;int num=0;for(int j=i;j>0;j=find(j-1)){num++;if(num*num>dp[i]) break;int next=find(j-1);dp[i]=min(dp[i],dp[next]+num*num);}}printf("%d\n",dp[n]);}return 0;}