[AcWing 899] 编辑距离

复杂度 $O(n \cdot m \cdot l^{2})$

总体复杂度 $1000 \times 1000 \times 10^{2} = 1 \times 10^{8}$

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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;
}

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1. 每次都要判断最短编辑距离是否小于等于所给的次数
2. $f[i][j] = min(f[i][j], f[i - 1][j - 1] + (a[i] \neq b[j]))$ 很巧妙