Zuma
time limit per test2 seconds
memory limit per test512 megabytes
inputstandard input
outputstandard 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.
題意：給你一串長度爲n的字符串，你每次可以刪除其中一段迴文串，問至少要刪除幾次才能完全刪除。
解題思路：對於區間[i,j],如果st[i]==st[j],那麼dp[i][j]==dp[i+1][j-1],因爲區間[i+1,j-1]一定能移除到一定程度與邊界2個構成迴文串,但如果區間長度等於2,dp[i][j]應該是等於1。而我們也能得到區間dp的轉移方程dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j])
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int dp[505][505],st[505];
int n;
int main(){
scanf("%d",&n);
for(int i=0;i<n;i++) scanf("%d",&st[i]);
for(int i=0;i<n;i++){
for(int j=i;j<n;j++) dp[i][j]=9999;
}
for(int i=0;i<n;i++){
dp[i][i]=1;
}
for(int l=2;l<=n;l++){
for(int i=0;i+l-1<n;i++){
int j=i+l-1;
if(st[i]==st[j]&&l>2) dp[i][j]=dp[i+1][j-1];
else if(st[i]==st[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]);
return 0;
}