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 AGTCTGACGC
11 AGTAAGTAGGC

Sample Output

4

Source

Manila 2006

又解决一道当年遗留下来的题……

这题类似最长公共子串，只是稍微变了一下罢了……注意初始化……

直接上代码，没什么好说的了……多校第一场貌似也有一道这种变题，这两天解决去……

#include <stdio.h>
#include <algorithm>
using namespace std;

#define INF 999999999

char str1[1005];
char str2[1005];
int dp[1005][1005];

int main()
{
    int i,j,n,m,x,y;
    while(scanf("%d",&n)!=EOF)
    {
        scanf("%s",str1);
        scanf("%d",&m);
        scanf("%s",str2);
        for (i=0;i<=n;i++)
        {
            dp[i][0]=i;
        }
        for (i=0;i<=m;i++)
        {
            dp[0][i]=i;
        }
        dp[0][0]=0;
        for (i=1;i<=n;i++)
        {
            for (j=1;j<=m;j++)
            {
                x=i-1;
                y=j-1;
                if (str1[x]==str2[y]) dp[i][j]=min(dp[i-1][j-1],min(dp[i-1][j],dp[i][j-1])+1);
                else dp[i][j]=min(min(dp[i-1][j],dp[i][j-1]),dp[i-1][j-1])+1;
            }
        }
        printf("%d\n",dp[n][m]);
    }
    return 0;
}