poj3356 AGTC 解题心得

原题:

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


分析:
dp[i][j]表示第一个字符串的前i个字符要和第二个字符串的前j个字符匹配需要的最少操作次数  
初始化dp[i][0]=i,若第二个字符串为0，只好把第一个字符串全删掉，所以dp[i][0]=i  同理，dp[0][j]=j  
状态转移方程：若s1[i-1]==s2[j-1],则不需要操作，那么dp[i][j]=dp[i-1][j-1] 
 否则，我们可能有三步操作，删除，插入，变换，所以dp[i][j]=min(dp[i-1][j]+1,dp[i][j-1]+1,dp[i-1][j-1]+1)  
把删除和插入看作是一个操作，自然dp[i-1][j-1]+1对应的是变换了。  

代码:

#include<iostream>
#include<cstdlib>
#include<stdio.h>
#include<memory.h>
using namespace std;
int dp[1010][1010];
char s1[1010];
char s2[1010];
int main()
{
    int l1,l2;
    while(scanf("%d%s%d%s",&l1,s1,&l2,s2)!=EOF)
    {
        for(int i=0;i<=l1;i++)
        for(int j=0;j<=l2;j++)
        dp[i][j]=0;
        for(int i=1;i<=l1;i++) dp[i][0]=i;
        for(int j=1;j<=l2;j++) dp[0][j]=j;
        for(int i=1;i<=l1;i++)
        for(int j=1;j<=l2;j++)
        {
            if(s1[i-1]==s2[j-1]) dp[i][j]=dp[i-1][j-1];
            else dp[i][j]=min(dp[i-1][j]+1,dp[i][j-1]+1),dp[i][j]=min(dp[i][j],dp[i-1][j-1]+1);
        }
        cout<<dp[l1][l2]<<endl;
    }
}