AcWing 902. 最短编辑距离

#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1010;
int n, m;
char a[N], b[N];
int f[N][N];//a的前N个去匹配b的前N个需要编辑的最小操作 
int main() {
    scanf("%d%s", &n, a + 1);
    scanf("%d%s", &m, b + 1);
    for (int i = 0; i <= m; i ++ ) f[0][i] = i;//a的前0个字母去匹配b的前i个字母，只能加 
    for (int i = 0; i <= n; i ++ ) f[i][0] = i;//a的前i个字母去匹配b的前0个字母，只能减 
    for (int i = 1; i <= n; i ++ )
        for (int j = 1; j <= m; j ++ ) {
//                      删除操作              增
            f[i][j] = min(f[i - 1][j] + 1, f[i][j - 1] + 1);
//                        改
            if (a[i] == b[j]) f[i][j] = min(f[i][j], f[i - 1][j - 1]);
            else f[i][j] = min(f[i][j], f[i - 1][j - 1] + 1);
        }
    printf("%d\n", f[n][m]);
    return 0;
}