LCS变形题

1、http://poj.org/problem?id=3356

2、题目大意：

给定两个字符串，目的是将第一串转换成和第二串相同的字符串，可以有三种操作，1、可以有增加、删除和改变的操作，删除即如果y中有，x没有，则在同位置y中可以删去，改变是指，x中的字符可以改成跟Y中的字符一样的，求得是最少几步可以将x字符串转换成y字符串

   此题类似于求最长公共子序列，状态转移方程为

                    if(a[i-1]==b[j-1])
                    dp[i][j]=minn(dp[i-1][j-1],dp[i-1][j]+1,dp[i][j-1]+1);
                else
                    dp[i][j]=minn(dp[i-1][j-1],dp[i-1][j],dp[i][j-1])+1;

题目：

AGTC

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7633		Accepted: 3037

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

 

4、代码：

#include<iostream>
using namespace std;
int minn(int x,int y,int z)
{
    if(x>y)
      x=y;
    if(x>z)
      x=z;
    return x;
}
int main()
{
    int n,m,i,j,dp[1100][1100];
    char a[1100],b[1100];

    while(cin>>n)
    {
    for(i=1;i<=n;i++)
      cin>>a[i];
       cin>>m;
    for(j=1;j<=m;j++)
       cin>>b[j];
    dp[0][0]=0;
    int n1=max(n,m);
    for(i=1;i<=n1;i++)
        dp[i][0]=i;
     for(i=1;i<=n1;i++)
        dp[0][i]=i;
     for(i=1;i<=n;i++)
     {
         for(j=1;j<=m;j++)
         {
             int t1,t2,t3;
              t2=dp[i-1][j]+1;//插入
              t3=dp[i][j-1]+1;//删除
             if(a[i]==b[j])
                 t1=dp[i-1][j-1];//修改
                 else
               t1=dp[i-1][j-1]+1;//修改
             dp[i][j]=minn(t1,t2,t3);

         }

     }
     cout<<dp[n][m]<<endl;
          }

     return 0;
}