POJ 3356 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:

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

LCS可以过，非LCS也可以过=-=

题意：由第一个序列到第二个序列最小操作步数。下面是两种方法。

非LCS：

#include
      
       
#include
       
         #include
        
          #include
         
           #include
          
            using namespace std; const int maxn=1100; char s1[maxn],s2[maxn]; int dp[maxn][maxn],len1,len2; int main() { while(~scanf("%d%s",&len1,s1)) { getchar(); scanf("%d%s",&len2,s2); for(int i=0;i<=len1;i++) dp[i][0]=i; for(int i=0;i<=len2;i++) dp[0][i]=i; for(int i=0;i
           
            
 
            
LCS:无法证明其正确性，discuss里看到的
 
            
 
            
#include
             
              
#include
              
                #include
               
                 #include
                
                  #include
                 
                   using namespace std; int len1,len2; int dp[1100][1100]; char s1[1100],s2[1100]; void LCS() { memset(dp,0,sizeof(dp)); for(int i=1;i<=len1;i++) { for(int j=1;j<=len2;j++) { if(s1[i-1]==s2[j-1]) dp[i][j]=dp[i-1][j-1]+1; else dp[i][j]=max(dp[i-1][j],dp[i][j-1]); } } } int main() { while(~scanf("%d%s",&len1,s1)) { getchar(); scanf("%d%s",&len2,s2); LCS(); cout<