题目链接：51nod 1183 编辑距离
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 using namespace std; 5 const int N = 1001; 6 char a[N], b[N]; 7 int dp[N][N];//dp[i][j]:a串的前i个字符转化成b串的前j个字符的最少操作数 8 int main(){ 9 int i, j; 10 scanf("%s%s", a+1, b+1); 11 int alen = strlen(a+1); 12 int blen = strlen(b+1); 13 for(i = 0; i <= alen; ++i) 14 dp[i][0] = i; 15 for(j = 0; j <= blen; ++j) 16 dp[0][j] = j; 17 for(i = 1; i <= alen; ++i) 18 for(j = 1; j <= blen; ++j) 19 dp[i][j] = min(dp[i-1][j-1] + (a[i]!=b[j]), min(dp[i-1][j]+1, dp[i][j-1]+1)); 20 printf("%d ", dp[alen][blen]); 21 return 0; 22 }