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:

• 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 AGTCTGACGC11 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<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<limits.h>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<len1;i++)        {            for(int j=0;j<len2;j++)            {                if(s1[i]==s2[j])                    dp[i+1][j+1]=min(dp[i][j],min(dp[i+1][j]+1,dp[i][j+1]+1));                else                    dp[i+1][j+1]=min(dp[i][j]+1,min(dp[i+1][j]+1,dp[i][j+1]+1));            }        }        printf("%d\n",dp[len1][len2]);    }    return 0;}

LCS:无法证明其正确性，discuss里看到的

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<limits.h>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<<max(len1,len2)-dp[len1][len2]<<endl;    }    return 0;}