UVA11584-Partitioning by Palindromes(动态规划基础)

Problem UVA11584-Partitioning by Palindromes

Accept: 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 1000 lowercase letters, with no whitespace within.

 Output

For each test case, output a line containing the minimum number of groups required to partition the input into groups of palindromes.

 

 Sample Input

3

racecar

fastcar

aaadbccb

 

Sample Output

1

7

3

题解：思路很明显，dp[i]的含义是前i个字符组成的字符串所能划分成的最少回文串的个数，定义好这个状态就很简单了，dp[i]肯定是从dp[j]转移过来（j<i）并且需要j+1到i是回文串，此时

dp[i] = min(dp[i], dp[j] + 1)，方程有了，边界也没啥困难的地方。

 #include <bits/stdc++.h>

 using namespace std;

 const int maxn =  + ;
 const int INF = 0x3f3f3f3f;

 char str[maxn];

 int is_palindromes[maxn][maxn];
 int dp[maxn];

 int Is_palindromes(int j, int i) {
     if (j >= i) return ;
     if (is_palindromes[j][i] != -) return is_palindromes[j][i];

     if (str[i] == str[j]) {
         return is_palindromes[j][i] = Is_palindromes(j + , i - );
     }
     else return is_palindromes[j][i] = ;
 }

 int main()
 {
     //freopen("input.txt", "r", stdin);
     int iCase;
     scanf("%d", &iCase);
     while (iCase--) {
         scanf("%s", str + );
         memset(is_palindromes, -, sizeof(is_palindromes));
         dp[] = ;
         int len = strlen(str + );
         for (int i = ; i <= len; i++) {
             dp[i] = i;
             for (int j = ; j < i; j++) {
                 if (Is_palindromes(j + , i)) dp[i] = min(dp[i], dp[j] + );
             }
         }
         printf("%d\n", dp[len]);
     }
     return ;
 }