POJ 3356【简单DP】

题目：AGTC

题意：

如题

解题思路:

看代码

View Code

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <string>
 4 #include <cstring>
 5 #include <algorithm>
 6 #include <vector>
 7 #include <map>
 8 
 9 using namespace std;
10 
11 const int MAX = 1000 + 10;
12 char X[MAX], Y[MAX];
13 int DP[MAX][MAX];
14 
15 inline int getMin(int a, int b, int c)
16 {
17     int min = a < b ? a : b;
18     return min < c ? min : c;
19 }
20 
21 int main()
22 {
23     freopen("in.txt","r",stdin);
24     int xLen, yLen;
25     while(scanf("%d%s%d%s", &xLen, &X[1], &yLen, &Y[1]) == 4){
26     for(int i = 1; i <= xLen; ++i)
27         DP[i][0] = i;
28     for(int j = 1; j <= yLen; ++j)
29         DP[0][j] = j;
30     DP[0][0] = 0;
31     for(int i = 1; i <= xLen; ++i)
32     {
33         for(int j = 1; j <= yLen; ++j)
34         {
35             DP[i][j] = getMin(DP[i - 1][j] + 1, DP[i][j - 1] + 1, DP[i - 1][j - 1] + (X[i] == Y[j] ? 0 : 1));
36         }
37     }
38     printf("%d\n", DP[xLen][yLen]);
39     }
40     return 0;
41 }

posted on 2011-09-25 19:19 Kenfly 阅读(346) 评论(0) 编辑收藏举报