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  #include  #include  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");	}	}