[AcWing 899] 编辑距离
作者:互联网
复杂度 \(O(n \cdot m \cdot l^{2})\)
总体复杂度 $ 1000 \times 1000 \times 10^{2} = 1 \times 10^{8} $
点击查看代码
#include<iostream>
#include<cstring>
using namespace std;
const int N = 15, M = 1010;
int n, m;
char str[M][N];
int f[N][N];
int solve(char a[], char b[])
{
int la = strlen(a + 1), lb = strlen(b + 1);
for (int i = 0; i <= lb; i ++) f[0][i] = i;
for (int i = 0; i <= la; i ++) f[i][0] = i;
for (int i = 1; i <= la; i ++)
for (int j = 1; j <= lb; j ++) {
f[i][j] = min(f[i - 1][j] + 1, f[i][j - 1] + 1);
f[i][j] = min(f[i][j], f[i - 1][j - 1] + (a[i] != b[j]));
}
return f[la][lb];
}
int main()
{
cin >> n >> m;
for (int i = 0; i < n; i ++) cin >> str[i] + 1;
while (m --) {
char s[N];
int t, res = 0;
cin >> s + 1 >> t;
for (int i = 0; i < n; i ++)
if (solve(str[i], s) <= t) res ++;
cout << res << endl;
}
return 0;
}
- 每次都要判断最短编辑距离是否小于等于所给的次数
- \(f[i][j] = min(f[i][j], f[i - 1][j - 1] + (a[i] \neq b[j]))\) 很巧妙
标签:int,899,复杂度,times,char,编辑,str,strlen,AcWing 来源: https://www.cnblogs.com/wKingYu/p/16306481.html