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 = 1000 + 10;
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 1;
    if (is_palindromes[j][i] != -1) return is_palindromes[j][i];

    if (str[i] == str[j]) {
        return is_palindromes[j][i] = Is_palindromes(j + 1, i - 1);
    }
    else return is_palindromes[j][i] = 0;
}

int main()
{
    //freopen("input.txt", "r", stdin);
    int iCase;
    scanf("%d", &iCase);
    while (iCase--) {
        scanf("%s", str + 1);
        memset(is_palindromes, -1, sizeof(is_palindromes));
        dp[0] = 0;
        int len = strlen(str + 1);
        for (int i = 1; i <= len; i++) {
            dp[i] = i;
            for (int j = 0; j < i; j++) {
                if (Is_palindromes(j + 1, i)) dp[i] = min(dp[i], dp[j] + 1);
            }
        }
        printf("%d
", dp[len]);
    }
    return 0;
}