Uva11584-partitioning by palindromes (Dynamic planning Basics)

problem uva11584-partitioning by Palindromesaccept:1326 submit:7151
Time limit:3000 MSec problem Description

Input

Input begins with the number N of test cases. Each test case consists of a single line of between 1 and lowercase letters, with no whitespace within.

Outputfor each test case, output a line containing the minimum number of groups required to partition the input into group  S of Palindromes. Sample INPUT3 racecar Fastcar AAADBCCB sample Output

1

7

3

Solution: The idea is obvious, dp[i] meaning is the first character of the string can be divided into the minimum number of palindrome, the definition of this state is very simple, dp[i] must be transferred from DP[J] (j<i) and need j+1 to I is a palindrome, at this time

Dp[i] = min (Dp[i], dp[j] + 1), the equation has, the boundary is not difficult place.

1#include <bits/stdc++.h>2 3 using namespacestd;4 5 Const intMAXN = ++Ten;6 Const intINF =0x3f3f3f3f;7 8 CharSTR[MAXN];9 Ten intIS_PALINDROMES[MAXN][MAXN]; One intDP[MAXN]; A  - intIs_palindromes (intJinti) { -     if(J >= i)return 1; the     if(Is_palindromes[j][i]! =-1)returnIs_palindromes[j][i]; -  -     if(Str[i] = =Str[j]) { -         returnIs_palindromes[j][i] = Is_palindromes (j +1I1); +     } -     Else returnIs_palindromes[j][i] =0; + } A  at intMain () - { -     //freopen ("Input.txt", "R", stdin); -     inticase; -scanf"%d", &icase); -      while(icase--) { inscanf"%s", str +1); -memset (Is_palindromes,-1,sizeof(Is_palindromes)); todp[0] =0; +         intlen = strlen (str +1); -          for(inti =1; I <= Len; i++) { theDp[i] =i; *              for(intj =0; J < I; J + +) { $                 if(Is_palindromes (j +1, i)) dp[i] = min (Dp[i], dp[j] +1);Panax Notoginseng             } -         } theprintf"%d\n", Dp[len]); +     } A     return 0; the}

Uva11584-partitioning by palindromes (Dynamic planning Basics)