CodeForces-607B Zuma

B. Zuma time limit per test 2 seconds memory limit per test 512 megabytes input standard input output standard output

Genos recently installed the game Zuma on his phone. In Zuma there exists a line of n gemstones, the i-th of which has color ci. The goal of the game is to destroy all the gemstones in the line as quickly as possible.

In one second, Genos is able to choose exactly one continuous substring of colored gemstones that is a palindrome and remove it from the line. After the substring is removed, the remaining gemstones shift to form a solid line again. What is the minimum number of seconds needed to destroy the entire line?

Let us remind, that the string (or substring) is called palindrome, if it reads same backwards or forward. In our case this means the color of the first gemstone is equal to the color of the last one, the color of the second gemstone is equal to the color of the next to last and so on.

Input

The first line of input contains a single integer n (1 ≤ n ≤ 500) — the number of gemstones.

The second line contains n space-separated integers, the i-th of which is ci (1 ≤ ci ≤ n) — the color of the i-th gemstone in a line.

Output

Print a single integer — the minimum number of seconds needed to destroy the entire line.

Examples input 3 1 2 1 output 1 input 3 1 2 3 output 3 input 7 1 4 4 2 3 2 1 output 2 Note

In the first sample, Genos can destroy the entire line in one second.

In the second sample, Genos can only destroy one gemstone at a time, so destroying three gemstones takes three seconds.

In the third sample, to achieve the optimal time of two seconds, destroy palindrome 4 4 first and then destroy palindrome 1 2 3 2 1.

题意:给出一个全部由数字组成的字符串,在这个字符串中如果有一个子串是回文串那么就可以移除,问至少要移除多少次, 才能把这个字符串全部移除 题解:区间dp,对于区间[i,j],如果a[i]==a[j],那么dp[i][j]==dp[i+1][j-1],因为区间[i+1,j-1]一定能移除到一定程 度与边界2个构成回文串,但是要注意,如果区间长度等于2时,dp[i][j]应该是等于1 并且有状态转移方程dp[i][i+len-1]=min(dp[i][i+len-1],dp[i][k]+dp[k+1][i+len-1]) (2<=len<=n,i<=k<i+len-1)

#include<cstdio> #include<algorithm> #include<string.h> #include<stdlib.h> #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 using namespace std; typedef long long LL; int a[505],dp[505][505]; int main(){ int n; // freopen("in.txt","r",stdin); while(~scanf("%d",&n)){ for(int i =0;i<n;i++) scanf("%d",&a[i]); memset(dp,0x3f,sizeof(dp)); for(int i=0;i<n;i++) dp[i][i]=1;//长度为1的时候,要消除它当然是需要1次 for(int l=2;l<=n;l++){ for(int i=0;i+l-1<n;i++){ int j=i+l-1; if(a[i]==a[j]&&l>2) dp[i][j]=dp[i+1][j-1];//如果区间起始和结束位置相等,则一定能和[i+1,j-1]构成回文串 else if(a[i]==a[j]) dp[i][j]=1; for(int k=i;k<j;k++){ dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]); } } } printf("%d\n",dp[0][n-1]); } }