poj 3356 AGTC

AGTC

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8136		Accepted: 3223

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

求最短编辑距离。类似LCS的求法。

AC代码：

#include <cstring>
#include <string>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cmath>
#include <vector>
#include <cstdlib>
#include <iostream>
#define max2(a,b) ((a) > (b) ? (a) : (b))
#define min2(a,b) ((a) < (b) ? (a) : (b))

using namespace std;

int dp[1005][1005];
int main()
{
    string a,b;
    int len1,len2;
    while(cin>>len1>>a>>len2>>b)
    {
    for(int i=0;i<=len1;i++)
        dp[i][0]=i;
    for(int i=0;i<=len2;i++)
        dp[0][i]=i;
    for(int i=1;i<=len1;i++)
    for(int j=1;j<=len2;j++)
    {
        if(a[i-1]==b[j-1])
        dp[i][j]=min2(dp[i-1][j-1],min2(dp[i][j-1]+1,dp[i-1][j]+1));
        else
        dp[i][j]=min2(dp[i-1][j-1]+1,min2(dp[i][j-1]+1,dp[i-1][j]+1));
    }
    cout<<dp[len1][len2]<<endl;
    }
    return 0;
}