Optimal Keypad

Description 
Optimus Mobiles produces mobile phones that support SMS messages. The Mobiles have a keypad of 12 keys, numbered 1 to 12. There is a character string assigned to each key. To type in the n-th character in the character string of a particular key, one should press the key n times. Optimus Mobiles wishes to solve the problem of assigning character strings to the keys such that for typing a random text out of a dictionary of common words, the average typing effort (i.e. the average number of keystrokes) is minimal. 

Figure 1
To be more precise, consider a set of characters {a, b, c,..., z, +, *, /, ?} printed on a label tape as in Fig. 2. We want to cut the tape into 12 pieces each containing one or more characters. The 12 labels are numbered 1 to 12 from left to right and will be assigned to the keypad keys in that order. 

Figure 2
You are to write a program to find the 11 cutting positions for a given dictionary of common words. The cutting positions should minimize the average number of keystrokes over all common words in the dictionary. Your output should be a string of 11 characters, where character i in this string is the first character of the (i+1)^th label.

Input 
The first line contains a single integer t (1 <= t <= 10), the number of test cases. Each test case starts with a line, containing an integer M (1 <= M <= 10000), the number of common words in the test case. In each M subsequent line, there is a common word. Each common word contains at most 30 characters from the alphabet {a, b, c,..., z, +, *, /, ?}.

Output 
The output contains one line per test case containing an optimal cut string. Obviously, there may be more than a single optimal cut string, so print the optimal cut string which is the smallest one in lexicographic order.

Sample Input 

2
2
hi
ok
5
hello
bye
how
when
who

Sample Output 

bcdefghijko
bcdefhlnowy

Source 

/**
题目大意：给出“abcd...z+* /?"的序列，要求分为12段，作为手机的12个按键上的字符，使得我们用手机输入单词时按键的次数最少，单词的数量是10000每个长度最大为30。 
思路：注意到输入的顺序可以打乱，即输入“ab”和输入“ba”花费是一样的，我们可以预处理一下，统计这30个字符出现的次数，现在我们要做的就是把这 30个字符分成12份。容易想到的方程是 dp[i][j] = min{ dp[i - 1][k - 1] + sum(k, j) }; dp[i][j]表示前j个字符分成i份，sum(k, j)表示第k个字符到第j个字符划分在同一个按键内的花费；最后记录一下路径。 
**/ 

#include <iostream> 
#include <cstring> 
#include <cstdio> 
using namespace std;

const int N = 32;

char alph[] = " abcdefghijklmnopqrstuvwxyz+*/?";
int flect[140];
int num[N];
int s[N][N];
int dp[N][N];
int ans[N];

void init()
{ 
	for (int i = 1; i <= 30; i++)
		flect[(int) alph[i]] = i;
} 
char str[N];

int main()
{ 
	//freopen("in.txt", "r", stdin);
	init();
	int n, T, i, j, k;

	scanf("%d", &T);
	while (T--)
	{
		scanf("%d", &n);
		memset(num, 0, sizeof(num));
		for (i = 0; i < n; i++)
		{
			scanf("%s", str);
			for (j = 0; str[j]; j++)
				num[flect[(int)str[j]]]++;
		}
		//dp
		memset(dp, 0x3f, sizeof(dp));
		int sum = 0;
		for (i = 1; i <= 19; i++)
		{
			sum += num[i] * i;
			dp[1][i] = sum;
			s[1][i] = 1;
		}
		for (i = 2; i <= 12; i++)
			for (j = i; j <= 30; j++)
			{
				for (k = i; k <= j; k++)
				{
					sum = 0;
					for (int h = k; h <= j; h++)
						sum += num[h] *(h - k + 1);
					if (dp[i - 1][k - 1] + sum < dp[i][j])
					{
						dp[i][j] = dp[i - 1][k - 1] + sum;
						s[i][j] = k;
					}
				}
			}
			int cnt = 0;
			int now = 30;
			for (i = 12; i > 1; i--)
			{
				ans[cnt++] = s[i][now];
				now = s[i][now] - 1;
			}
			for (i = cnt - 1; i >= 0; i--)
				printf("%c", alph[ans[i]]);
			printf("\n");
	}
	
}