【动态规划】字符串距离

 

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int N=3000;
 4 char c1[N],c2[N];
 5 int l1,l2,d[N][N],k;
 6 int main() {
 7     scanf("%s", c1);
 8     scanf("%s", c2);
 9     l1 = strlen(c1);
10     l2 = strlen(c2);
11     scanf("%d", &k);
12     for (int i = 1; i <= l1; i++) {
13         d[i][0] = d[i - 1][0] + k;
14     }
15     for (int i = 1; i <= l2; i++) {
16         d[0][i] = d[0][i - 1] + k;
17     }
18     for (int i = 1; i <= l1; i++) {
19         for (int j = 1; j <= l2; j++) {
20             d[i][j] = min(d[i - 1][j] + k, d[i][j - 1] + k);
21             d[i][j] = min(d[i][j], d[i - 1][j - 1] + abs(c1[i - 1] - c2[j - 1]));
22         }
23     }
24     printf("%d\n", d[l1][l2]);
25 }
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posted @ 2019-07-23 19:03  Snow_in_winer  阅读(367)  评论(0)    收藏  举报