acwing 899 编辑距离 （线性DP）

题面

题解

1. 最短编辑距离的应用，我们只需要遍历每个字符串，求每个字符串和输入的字符串的最短距离是否小于等于限制，如果符合答案就+1

2. 

代码

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>

using namespace std;
const int N = 1100;

int n, m;
char str[N][N];
int f[N][N];

int editor(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);
            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);
        }
    }

    return f[la][lb];
}

int main() {
    
    

    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);

    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> str[i] + 1;

    while (m--) {
    
    
        char s[N];
        int limit;
        cin >> s + 1 >> limit;
        int res = 0;
        for (int i = 1; i <= n; i++) {
    
    
            if (editor(s, str[i]) <= limit) {
    
    
                res++;
            }
        }
        cout << res << endl;
    }
    return 0;
}