POJ 3356 AGTC（dp之公共最长子序列）

最新推荐文章于 2019-12-12 19:48:37 发布

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#dp #公共最长子序列

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本文介绍了一种通过寻找两个字符串间的最长公共子序列来计算从一个字符串转换到另一个字符串所需的最少操作数的方法，并提供了一个实现该算法的C++示例程序。

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Let x and y be two strings over some finite alphabet A. We would like to transform xinto 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

题解：

很简单的一题，就是找公共最长子序列，然后用最长的串减去公共子序列长度就是答案。。练习一下公共最长子序列

代码：

#include<iostream>
#include<stdio.h>
#include<map>
#include<algorithm>
#include<math.h>
#include<queue>
#include<stack>
#include<string>
#include<cstring>
using namespace std;
char s1[1005],s2[1005];
int dp[1005][1005];
int main()
{
    int i,j,k,n,m;
    while(scanf("%d%s%d%s",&n,s1,&m,s2)!=EOF)
    {
        dp[0][0]=0;
        dp[0][1]=0;
        dp[1][0]=0;
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=m;j++)
            {
                if(s1[i-1]==s2[j-1])
                {
                    dp[i][j]=dp[i-1][j-1]+1;
                }
                else
                {
                    dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
                }
            }
        }
        printf("%d\n",max(n,m)-dp[n][m]);
    }
    return 0;
}