poj1732

题目大意：给你一个a..z与数字的对应表及一些单词

               再给你一个目标串数字串。。问你最少用那些单词可以拼成数字串。。

思路：dp

        f[i]表示前i个最少用几个单词拼成

       f[i]= f[i - len[k]] + 1     (第k个单词对应数字正好相符)

/*
  State:Accepted
  Time:2013-03-10 14:35:57
*/
#include <cstring>
#include <string>
#include <fstream>
#include <cstdlib>
#include <cstdio>
using namespace std;
char s[150],s1[50001][55];
const int num[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};
int m , a[50010][55] , f[200] , fpt[200],n;
void init(){
    scanf("%s",&s);
    n = strlen(s); 
    scanf("%d",&m);
    for (int i = 1; i <= m; ++i){
          scanf("%s",&s1[i]);
          a[i][0] = strlen(s1[i]);
          for (int j = 0; j < strlen(s1[i]); ++j)
             a[i][j + 1] = num[s1[i][j] - 'a'];
    }
}

void print(int now){
     if (now <= 0) return;
     int len = a[fpt[now]][0];
     print(now - len);
    // printf("ID=%d J=%d :",now , fpt[now]);
     printf("%s ",s1[fpt[now]]); 
}

void dp(){
     bool bo;
     int now;
     for (int i = 0; i <= n; ++i)  
         f[i] = 0xffffff;
     f[0] = 0;
     for  (int i = 0; i < n; ++i)
          for (int j = 1; j <= m; ++j)
           {
              bo = true;
              for (int l = 1 ; l <= a[j][0]; ++l)
                 if  (i + l > n || a[j][l] != num[ s[i + l] - 'a'])
                   {
                      bo = false;
                      break;
                   };
              now = i + a[j][0];
              if (bo && f[i] + 1 < f[now]){
                       f[now] =  f[i] + 1;
                       fpt[now] = j;
              }
            }
     print(n);
     printf("\n");
}

int main(){
    freopen("poj1732.in","r",stdin);
    freopen("poj1732.out","w",stdout);
    init();
    dp();
}