hdu 5009 Paint Pearls 2014 ACM/ICPC Asia Regional Xi'an Online

Paint Pearls

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1014    Accepted Submission(s): 325

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.

 

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,...,a_n (1 ≤ a_i ≤ 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

 

Source

2014 ACM/ICPC Asia Regional Xi'an Online

 

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題意：區間塗色，花費爲顏色數的平方，求最小的花費。

思路：思路直接看代碼。比賽的時候想到過dp，但是o(n^2)的算法必定超時 ，就沒有做，比完看居然n^2的算法加剪枝就可以過掉。g++1.4s

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
const int N = 50005;
const int inf=0x3f3f3f3f;
struct node
{
    int id,c,l;
}t[N];
int c,n,cnt;
int vis[N],dp[N];
vector<int> v;
bool cmp1(node a,node b)
{
    return a.c<b.c;
}
bool cmp2(node a,node b)
{
    return a.id<b.id;
}
void read_int(int&x)
{
    char ch=getchar();
    while(ch<'0'||ch>'9') ch=getchar();
    x = ch - '0';
    ch=getchar();
    while('0'<=ch&&ch<='9')
    {
        x = 10*x + ch-'0';
        ch = getchar();
    }
}
int main(){
    while(scanf("%d", &n)!=EOF){
        read_int(t[1].c);
        int id=1;
        t[1].id=1;
        for(int i=2;i<=n;i++){
            read_int(c);
            if(c!=t[id].c){
                t[++id].c=c;
                t[id].id=id;
            }
        }
        n=id;
        sort(t+1,t+n+1,cmp1);
        id=0;
        t[1].l=0;
        for(int i=2;i<=n;i++){
            if(t[i].c!=t[i-1].c)id++;
            t[i].l=id;
        }
        sort(t+1,t+n+1,cmp2);
        memset(dp,inf,sizeof(dp));
        memset(vis,0,sizeof(vis));
        dp[0]=0;
        dp[n]=n;
        for(int i=0;i<n;i++){
            cnt=0;
            if(dp[i]>=dp[i+1])continue;
            for(int j=i+1;j<=n;j++){
                if(!vis[t[j].l]){
                    v.push_back(t[j].l);
                    vis[t[j].l]++;
                    cnt++;
                }
                if(dp[i]+cnt*cnt>=dp[n])break;
                dp[j]=min(dp[j],dp[i]+cnt*cnt);
            }
            for(int j=0;j<v.size();j++)
                vis[v[j]]=0;
            v.clear();
        }
        printf("%d\n",dp[n]);
    }
    return 0;
}