POJ 3356(最短编辑距离问题)

POJ - 3356

AGTC

Time Limit: 1000MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

 

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

题解：给你两行字符串，第一行为A长度m，第二行为B，长度n，dp[i][j]表示编辑A前i字母和编辑B的前j个字母所需要的操作次数。这里我们要求最少的操作次数。

  我们可以执行的操作有删除，插入，还有替换。

解题思路：

长当某一个字符串为空时，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。

如果A比B：

如果没有空字符串，有3中操作选择，
1.替换 ，将a[0]和b[0]判断，如果相等，即返回上一个字符的，操作次数加1，即dp[i-1][j-1]+1.
2.删除，所以删除A一个字符，
也就是必须有一次操作，删除A[ ]后，返回A的前一个字符，即dp[i-1][j]+1
3.插入，在B添加一个字符，A不变，如果执行完相等，就要找B的前一个去匹配，所以要减1，和A去匹配，执行了一次，操作次数要加1，即dp[i][j-1] + 1

如果B比A长，则删除和插入交换。

每次选出这三种执行操作数最少的就可以了。

下面的代码

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int dp[1005][1005];
char a[1005],b[1005];
int m,n;
int main()
{
     while(~scanf("%d %s",&m,a+1))
     {
         scanf("%d %s",&n,b+1);
         int maxl=max(m,n);
         for(int i=0;i<=maxl;i++)
         {
             dp[i][0]=i;
             dp[0][i]=i;
         }
         for(int i=1;i<=m;i++)
        {
            for(int j=1;j<=n;j++)
         {
             if(a[i]==b[j])
                dp[i][j]=dp[i-1][j-1];
             else
                dp[i][j]=min(min(dp[i-1][j]+1,dp[i][j-1]+1),dp[i-1][j-1]+1);
         }
        }
         cout<<dp[m][n]<<endl;
     }
}