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。

 1 #include <bits/stdc++.h>
 2
 3 using namespace std;
 4
 5 const int maxn = 1000 + 10;
 6 const int INF = 0x3f3f3f3f;
 7
 8 char str[maxn];
 9
10 int is_palindromes[maxn][maxn];
11 int dp[maxn];
12
13 int Is_palindromes(int j, int i) {
14     if (j >= i) return 1;
15     if (is_palindromes[j][i] != -1) return is_palindromes[j][i];
16
17     if (str[i] == str[j]) {
18         return is_palindromes[j][i] = Is_palindromes(j + 1, i - 1);
19     }
20     else return is_palindromes[j][i] = 0;
21 }
22
23 int main()
24 {
25     //freopen("input.txt", "r", stdin);
26     int iCase;
27     scanf("%d", &iCase);
28     while (iCase--) {
29         scanf("%s", str + 1);
30         memset(is_palindromes, -1, sizeof(is_palindromes));
31         dp[0] = 0;
32         int len = strlen(str + 1);
33         for (int i = 1; i <= len; i++) {
34             dp[i] = i;
35             for (int j = 0; j < i; j++) {
36                 if (Is_palindromes(j + 1, i)) dp[i] = min(dp[i], dp[j] + 1);
37             }
38         }
39         printf("%d\n", dp[len]);
40     }
41     return 0;
42 }

原文地址：https://www.cnblogs.com/npugen/p/9746263.html