leetcode1278 Palindrome Partitioning III

思路：

使用动态规划，dp[i][j]表示将子串s[0..i]划分成j个回文串所需要的最小修改次数。

实现：

 1 class Solution
 2 {
 3 public:
 4     int palindromePartition(string s, int k)
 5     {
 6         int n = s.length();
 7         vectorint>> m(n, vector<int>(n, 0x3f3f3f3f));
 8         for (int i = n - 1; i >= 0; i--)
 9         {
10             for (int j = i; j < n; j++)
11             {
12                 if (j == i) m[i][j] = 0;
13                 else if (j == i + 1) m[i][j] = s[i] != s[j];
14                 else
15                 {
16                     m[i][j] = m[i + 1][j - 1];
17                     if (s[i] != s[j]) m[i][j]++;
18                 }
19             }
20         }
21         vectorint>> dp(n, vector<int>(k + 1, 0x3f3f3f3f));
22         for (int i = 0; i < n; i++)
23         {
24             dp[i][1] = m[0][i];
25             for (int p = 2; p <= min(k, i + 1); p++)
26             {
27                 for (int j = 0; j < i; j++)
28                 {
29                     dp[i][p] = min(dp[i][p], dp[j][p - 1] + m[j + 1][i]);
30                 }
31             }
32         }
33         return dp[n - 1][k];
34     }
35 }