CodeForces-607B Zuma（dp）

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：题目给出一个仅由数字构成的字符串，每次都可以去掉其中的一个回文子串，问移除整个字符串需要最少几次操作。

思路：一个经典的区间dp，dp[i][j]代表完全删除i到j这个区间需要最少几次操作。如果a[i-1]==a[j+1]，那么dp[i][j]必等于dp[i-1][j+1]; 否则dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j]);

具体看代码，挺好理解的。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<list>
#include<math.h>
#include<vector>
#include<stack>
#include<string>
#include<set>
#include<map>
#include<numeric>
#include<stdio.h>
#include<functional>
#include<time.h>
#pragma warning(disable:6031)

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 1e5 + 10;
const ll mode = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const double pi = 3.14159265358979323846264338327950;
template <class T> inline T min(T a, T b, T c) { return min(min(a, b), c); }
template <class T> inline T max(T a, T b, T c) { return max(max(a, b), c); }
template <class T> inline T min(T a, T b, T c, T d) { return min(min(a, b), min(c, d)); }
template <class T> inline T max(T a, T b, T c, T d) { return max(max(a, b), max(c, d)); }

int a[555], dp[555][555];

int main()
{
	int n;
	cin >> n;
	for (int i = 0; i < n; i++)
		cin >> a[i];
	memset(dp, inf, sizeof(dp));
	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 (a[i] == a[j] && l > 2)
				dp[i][j] = dp[i + 1][j - 1];
			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]);

	return 0;
}