POJ 1732 Phone numbers

 

Phone numbers

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2456		Accepted: 852		Special Judge

Description

In the present world you frequently meet a lot of call numbers and they are going to be longer and longer. You need to member such a kind of numbers. One method how to do it in an easy way is to assign letters to digits as shown in the following picture: 

	1 ij	2 abc	3 def

	4 gh	5 kl	6 mn

	7 prs	8 tuv	9 wxy

		0 oqz

This way every word or a group of words can be assigned a unique number, so you can remember words instead of call numbers. It is evident that it has its own charm if it is possible to find some simple relationship between the word and the person itself. So you can earn that the call number 941837296 of a chess playing friend of yours can be read as WHITEPAWN, and the call number 2855304 of your favourite teacher is read BULLDOG. 

Write a program to find the shortest sequence of words (i.e. one having the smallest possible number of words) which corresponds to a given number and a given list of words. The correspondence is described by the picture above.

Input

The first line of input contains the call number, the transcription of which you have to find. The number consists of at most 100 digits. The second line contains the total number of the words in the dictionary (maximum is 50000). Each of the remaining lines contains one word, which consists of maximally 50 small letters of the English alphabet. The total size of the input doesn't exceed 300KB.

Output

The only line of output contains the shortest sequence of words which has been found by your program. The words are separated by single spaces. If there is no solution to the input data, the line contains text `No solution.'. If there are more solutions having the minimum number of words, you can choose any single one of them.

Sample Input

7325189087
5
it
your
reality
real
our

Sample Output

reality our

Source

CEOI 1999

 

/* DP,类似矩阵链乘法，dps[i][j].minNum表示输入电话串中第i位至第j位子串需要的最少单词数， 则dps[i][j].minNum = min(dps[i][j].minNum, dps[i][k].minNum + dps[k + 1][j].minNum), i <= k < j dps[i][j].curPos记录这个最优的k 首先要做的是边输入单词边处理，对每一个单词在电话号码串中寻找匹配串，加入对于单词str在号码串中找 到匹配的字串str(i, j)则将dps[i][j].minNum 设置为1，dps[i][j].str = str */ #include <iostream> #include <string> #define minv(a, b) ((a) <= (b) ? (a) : (b)) #define maxv(a, b) ((a) >= (b) ? (a) : (b)) #define MAX_N 105 #define MAX_VAL 1000000 using namespace std; struct dp { int minNum; int cutPos; string str; }dps[MAX_N + 1][MAX_N + 1]; int len, wordNum, minLen; char map[26] = {'2', '2', '2', '3', '3', '3', '4', '4', '1', '1', '5', '5', '6', '6', '0', '7', '0', '7', '7', '8', '8', '8', '9', '9', '9', '0'}; string phone; void init() { int i, j; for(i = 0; i < len; i++) { for(j = 0; j < len; j++) { dps[i][j].minNum = MAX_VAL; dps[i][j].str = ""; dps[i][j].cutPos = -1; } } minLen = MAX_VAL; } //将字母单词word转化为数字序列digits void getDigits(const string &word, string &digits) { int len = word.length(); digits = word; for(int i = 0; i < len; i++) digits[i] = map[word[i] - 'a']; } void dp() { int l, f, t, k; //遍历长度 for(l = minLen; l <= len; l++) { //遍历起始处 for(f = 0; f <= len - l; f++) { t = f + l - 1; int curMinLen = MAX_VAL; int curMinPos = 0; //遍历中间拆分位置 for(k = f; k < t; k++) { if(curMinLen > dps[f][k].minNum + dps[k + 1][t].minNum) { curMinLen = dps[f][k].minNum + dps[k + 1][t].minNum; curMinPos = k; } } if(dps[f][t].minNum > curMinLen) { dps[f][t].minNum = curMinLen; dps[f][t].cutPos = curMinPos; } } } } //递归打印结果 void printRes(int s, int t) { if(dps[s][t].cutPos == -1) cout<<dps[s][t].str<<" "; else { printRes(s, dps[s][t].cutPos); printRes(dps[s][t].cutPos + 1, t); } } int main() { int i, f, t; string word; string digits; cin>>phone; len = phone.length(); init(); cin>>wordNum; for(i = 1; i <= wordNum; i++) { cin>>word; int wordLen = word.length(); if(wordLen < minLen) minLen = wordLen; getDigits(word, digits); for(f = 0; f <= len - wordLen; f++) { t = f + wordLen - 1; if(dps[f][t].minNum != MAX_VAL) continue; if(phone.substr(f, wordLen) == digits) { dps[f][t].minNum = 1; dps[f][t].str = word; } } } dp(); if(dps[0][len - 1].minNum == MAX_VAL) cout<<"No solution."<<endl; else { printRes(0, len - 1); printf("/n"); } return 0; }