codeforces 607B 区间dp

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.

题目链接

题意：给你一串长度为n的字符串，你每次可以删除其中一段回文串，问至少要删除几次才能完全删除。

解题思路：对于区间[i,j],如果st[i]==st[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,但如果区间长度等于2,dp[i][j]应该是等于1。而我们也能得到区间dp的转移方程dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j])

#include
#include
#include
#include
using namespace std;
int dp[505][505],st[505];
int n;
int main(){
    scanf("%d",&n);
    for(int i=0;iscanf("%d",&st[i]);
    for(int i=0;ifor(int j=i;j9999;
    }
    for(int i=0;i1;
    }
    for(int l=2;l<=n;l++){
        for(int i=0;i+l-1int 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;k1][j]);
            }
        }
    }
    printf("%d\n",dp[0][n-1]);
    return 0;
}