f[i][j] 表示第一串前 i 个到第二串前 j 个的最小编辑距离
f[i][j] = f[i - 1][j - 1] (s1[i] == s2[j])
f[i][j] = min(f[i - 1][j], f[i][j - 1], f[i - 1][j - 1]) (s1[i] != s2[j])
边界 f[0][j] = j (1 <= j <= m)
f[i][0] = i (1 <= i <= n)
——代码
1 #include <cstdio> 2 #include <cstring> 3 4 int n, m; 5 int f[2048][2048]; 6 char s1[2048], s2[2048]; 7 8 inline int min(int x, int y) 9 { 10 return x < y ? x : y; 11 } 12 13 int main() 14 { 15 int i, j; 16 scanf("%s %s", s1 + 1, s2 + 1); 17 n = strlen(s1 + 1); 18 m = strlen(s2 + 1); 19 for(i = 1; i <= n; i++) f[i][0] = i; 20 for(i = 1; i <= m; i++) f[0][i] = i; 21 for(i = 1; i <= n; i++) 22 for(j = 1; j <= m; j++) 23 { 24 if(s1[i] == s2[j]) f[i][j] = f[i - 1][j - 1]; 25 else f[i][j] = min(min(f[i - 1][j], f[i][j - 1]), f[i - 1][j - 1]) + 1; 26 } 27 printf("%d ", f[n][m]); 28 return 0; 29 }