POJ 3356 AGTC

Posted 2020-09-05 谦谦君子，陌上其华

http://poj.org/problem?id=3356

题意：

给出两个字符串，通过转换，添加，删除使两个字符串相等，求最少操作次数。

 

思路：

d[i][j]表示第一个字符串取到第i位，第二个字符串取到第j位时的最少操作次数。

 1 #include<iostream>
 2 #include<string>
 3 #include<cstring>
 4 #include<cstdio>
 5 #include<algorithm>
 6 using namespace std;
 7 
 8 const int maxn = 1000 + 5;
 9 
10 int n1, n2;
11 char s1[maxn], s2[maxn];
12 int d[maxn][maxn];
13 
14 int main()
15 {
16     //freopen("D:\\txt.txt", "r", stdin);
17     while (cin >> n1)
18     {
19         cin >> s1 + 1 >> n2 >> s2 + 1;
20         memset(d, 0, sizeof(d));
21         for (int i = 1; i <= n1; i++)
22             d[i][0] = i;
23         for (int i = 1; i <= n2; i++)
24             d[0][i] = i;
25         for (int i = 1; i <= n1; i++)
26         for (int j = 1; j <= n2; j++)
27         {
28             if (s1[i] == s2[j])  d[i][j] = d[i - 1][j - 1];
29             else
30                 d[i][j] = min(d[i - 1][j] + 1, min(d[i][j - 1] + 1, d[i - 1][j - 1] + 1));
31         }
32         cout << d[n1][n2] << endl;
33     }
34 }