[codeforces1132F]Clear the String

time limit per test : 3 seconds
memory limit per test : 256 megabytes

You are given a string $s$ of length $n$ consisting of lowercase Latin letters. You may apply some operations to this string: in one operation you can delete some contiguous substring of this string, if all letters in the substring you delete are equal. For example, after deleting substring $bbbb$ from string $abbbbaccdd$ we get the string $aaccdd$.
Calculate the minimum number of operations to delete the whole string $s$.

Input

The first line contains one integer $n(1\le n\le 500)$ — the length of string $s$.
The second line contains the string $s(|s|=n)$ consisting of lowercase Latin letters.

Output

Output a single integer — the minimal number of operation to delete string $s$ .

题意：
给一个字符串，你每次操作可以删除一段连续的相同字符，例如字符串$abbbbaccdd$删去$bbbb$之后变成$accdd$，请问至少操作多少次才会将这个字符串删没。

题解：
$f[i][j][k]$表示为把$i$到$j$这个区间删到只剩下字母$k$的最少操作。
$f[i][j][a]=min(f[i][j][a],f[i][k][a]+f[k+1][j][b]+(a!=b))$
这个算法复杂度是$O((n^3)\ast 26\ast 26)$
但是其实关于字母的状态数比较少，可以用map优化这个算法，将$f[i][j]$设为map,然后只要枚举存在的状态即可。

#include
#define LiangJiaJun main
#define INF n+4
using namespace std;
int n;
map<int,int>f[504][504];
char s[504];
int LiangJiaJun(){
	scanf("%d",&n);
	scanf("%s",s+1);
	for(int l=1;l<=n;l++){
		for(int i=1;i+l-1<=n;i++){
			int j=i+l-1;
			if(l==1)f[i][j][s[i]-'a']=0;
			for(int k=i;k<j;k++){
				for(auto a:f[i][k]){
					for(auto b:f[k+1][j]){
						if(!f[i][j].count(a.first))f[i][j][a.first]=INF;
						f[i][j][a.first]=min(f[i][j][a.first],(a.second)+(b.second)+((a.first)!=(b.first)));
					}
				}
			}
		}
	}
	int mint=INF;
	for(int i=0;i<26;i++){
        if(!f[1][n].count(i))continue;
        mint=min(mint,f[1][n][i]);
	}
	printf("%d\n",mint+1);
	return 0;
}