emmm...好難的一道dp

其實本質是兩次dp.第一次求出都不相等時候的最優解, 第二次在求出兩串有相同時的更優的解.

https://blog.csdn.net/hndu__lz/article/details/52167981

這位講的不錯, 可以去看看.

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 #include <algorithm>
 5 using namespace std;
 6 const int MAXN = 1e2 + 20;
 7 
 8
 
 char s[MAXN], t[MAXN];
 9 int f[MAXN][MAXN], g[MAXN];
10 int N;
11 
12 int main()
13 {
14     while(scanf("%s\n%s", s, t) == 2)
15     {
16         N = strlen(s);
17         for(int i = 0; i < N; i++) f[i][i] = 1;
18             
19         for(int len = 1; len < N; len++)
20             for(int
 
 l = 0; l + len < N; l++){
21                 int r = l + len;
22                 f[l][r] = f[l + 1][r] + 1;
23                 
24                 for(int k = l + 1; k <= r; k++)
25                     if(t[l] == t[k]) f[l][r] = min(f[l][r], f[l + 1][k] + f[k + 1][r]);
26             }
27             
28
 
         for(int r = 0; r < N; r++)
29         {
30             g[r] = f[0][r];
31             if(s[r] == t[r]) g[r] = (r == 0) ? 0 : g[r - 1];
32             else 
33                 for(int l = 0; l < r; l++) g[r] = min(g[r], g[l] + f[l + 1][r]);
34         }
35         cout<<g[N - 1]<<endl;
36     }
37     return 0;
38 }

UVa1437 String painter