poj4323 最短编辑距离

AGTC

Description

Let x and y be two strings over some finite alphabet A. We would like to transform x into y allowing only operations given below:

• Deletion: a letter in x is missing in y at a corresponding position.
• Insertion: a letter in y is missing in x at a corresponding position.
• Change: letters at corresponding positions are distinct

Certainly, we would like to minimize the number of all possible operations.

Illustration

A G T A A G T * A G G C

| | |       |   |   | |

A G T * C * T G A C G C

Deletion: * in the bottom line
Insertion: * in the top line
Change: when the letters at the top and bottom are distinct

This tells us that to transform x = AGTCTGACGC into y = AGTAAGTAGGC we would be required to perform 5 operations (2 changes, 2 deletions and 1 insertion). If we want to minimize the number operations, we should do it like

A  G  T  A  A  G  T  A  G  G  C

|  |  |        |     |     |  |

A  G  T  C  T  G  *  A  C  G  C

and 4 moves would be required (3 changes and 1 deletion).

In this problem we would always consider strings x and y to be fixed, such that the number of letters in x is m and the number of letters in y is n where n ≥ m.

Assign 1 as the cost of an operation performed. Otherwise, assign 0 if there is no operation performed.

Write a program that would minimize the number of possible operations to transform any string x into a string y.

Input

The input consists of the strings x and y prefixed by their respective lengths, which are within 1000.

Output

An integer representing the minimum number of possible operations to transform any string x into a string y.

Sample Input

10 AGTCTGACGC
11 AGTAAGTAGGC

Sample Output

4

Source

Manila 2006

这道题是典型的DP问题最短编辑距离。

其状态转移方程为

1. #include <stdio.h>
2. #define MAXN 1024
3. int dp[MAXN][MAXN];
4. char str1[MAXN],str2[MAXN];
5. int min3(int a,int b,int c){
6. int min=a;
7. if(min>b) min=b;
8. if(min>c) min=c;
9. return min;
10. }
11. int main()
12. {
13. int n,m;
14. while(scanf("%d%s",&n,str1)!=EOF){
15. scanf("%d%s",&m,str2);
16. for(int i=0;i<=n;i++)
17. dp[i][0]=i;
18. for(int j=0;j<=m;j++)
19. dp[0][j]=j;
20. int count;
21. for(int i=1;i<=n;i++){
22. for(int j=1;j<=m;j++){
23. str1[i-1]==str2[j-1]?count=0:count=1;
24. dp[i][j]=min3(dp[i-1][j-1]+count,dp[i-1][j]+1,dp[i][j-1]+1);
25. }
26. }
27. printf("%d\n",dp[n][m]);
28. }
29. return 0;
30. }

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