POJ3356 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



题意：  给出两个字符串x 与 y，其中x的长度为n，y的长度为m，并且m>=n
　　　　然后y可以经过删除一个字母，添加一个字母，转换一个字母，三种操作得到x
　　　　问最少可以经过多少次操作


解题思路：（转至 贰圣的博客）

　　　　我们设dp[i][j]的意义为y取前i个字母和x取前j个字母的最少操作次数

那么可以得到dp[0][i] = i和dp[i][0]=i，因为某一字符串为空的，要得到另一个i长度字符串，必须经过i次插入操作。 而dp[1][1]，有3种操作 　　　　　　 　　　　1.转换 ，将str1[0]和str2[0]判断，如果相等，则dp[1][1]=0，否则dp[1][1]=1

2.删除，因为，目的串比源串小，所以删除源串一个字符， 也就是必须有一次操作，删除str1[0]后，那么dp[1][1]就是dp[0][1]的值+1

3.添加，在目的串添加一个字符，即源串不变，但是目的串减1，和源串去匹配，即dp[1][0] + 1

这样dp[i][j]可以得到3中操作的最小值

dp[i-1][j-1]+(str1[i]==str2[j]?0:1)

dp[i-1][j]+1

dp[i][j-1]+1

 

 

代码如下：

 

 

#include <stdio.h>
#include <string.h>
char x[10050],y[10050];
int d[10050][10050];
int min(int a,int b)
{
    return a<b?a:b;
}
int main()
{
    int n,m;
    while(scanf("%d%s",&n,&x)==2)
    {
        scanf("%d%s",&m,&y);
        d[0][0];
        for(int i=1;i<=n;i++)
            d[i][0]=i;
        for(int i=1;i<=m;i++)
            d[0][i]=i;
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
        {
            int tmp=d[i-1][j-1]+(x[i-1]==y[j-1]?0:1);
            tmp=min(d[i][j-1]+1,tmp);
            d[i][j]=min(d[i-1][j]+1,tmp);
        }
        printf("%d\n",d[n][m]);
    }
    return 0;
}