poj4323 最短编辑距离

Posted 2020-09-30 第三方as

AGTC

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12240		Accepted: 4594

 

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.    
12. int main()  
13. {  
14.      int n,m;  
15.       while(scanf("%d%s",&n,str1)!=EOF){  
16.         scanf("%d%s",&m,str2);  
17.         for(int i=0;i<=n;i++)  
18.             dp[i][0]=i;  
19.         for(int j=0;j<=m;j++)  
20.             dp[0][j]=j;  
21.               
22.         int count;  
23.         for(int i=1;i<=n;i++){  
24.             for(int j=1;j<=m;j++){  
25.                 str1[i-1]==str2[j-1]?count=0:count=1;  
26.                 dp[i][j]=min3(dp[i-1][j-1]+count,dp[i-1][j]+1,dp[i][j-1]+1);  
27.             }  
28.         }  
29.         printf("%d\n",dp[n][m]);  
30.     }  
31.     return 0;  
32. }  

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