POJ3356 AGTC （最短编辑距离问题）

AGTC

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7923		Accepted: 3142

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

Source

Manila 2006

 

题意：

由字符串 s1 通过下列三种操作

　1、插入一个字符；

   2、删除一个字符；

　3、改变一个字符

变换到字符串 s2 所需要的最少操作次数（亦即最短编辑距离问题）

解析：

第三届软件大赛决赛中关于核苷酸 变换的问题解析http://blog.csdn.net/kuaisuzhuceh/article/details/8680799如下（同样也适用于该题）；
状态转移方程：
有三种情况可以导致我们上面设计的状态会发生转移。我们现在来看A[i] 和 B[j] ，
①、我们可以在 B[j]后面插入一个核苷酸（即一个字符）ch，ch==A[i]，这样做的话，
至少需要 dp[i - 1][j] + 1步操作，即 dp[i][j] = dp[i - 1][j] + 1。
②、我们可以删除 B[j]，这样的话，B[1...j] 变为A[1...i] 需要 dp[i][j - 1]步，
即 dp[i][j] = dp[i][j - 1] + 1。
③、我们也可以考虑修改 B[j]，使它变为A[j]，但是如果 B[j]本来就等于 A[i]的话，
那修改其实相当于用了 0步，如果 B[j] != A[i] 的话，那修改相当于用了 1步。
所以 dp[i][j] = dp[i - 1][j - 1] + （A[i] == B[j] ? 0, 1）。

决策：
决策就很简单了，从上面三种状态转移中选择一个最小值就可以了。

处理边界：
处理好边界非常重要，这里需要注意的是对dp[0][0....m],dp[0.....n][0]的初始化，
可以这样看，dp[0][i],就是说A[1...n]是一个空串，而B[1...m]十个长度为i的串，
很显然B串变为A串就是删除i个核苷酸。

#include <cstdio>
#include <iostream>
#include <cstring>

using namespace std;

int len1, len2;
int dp[1005][1005];//  dp[i][j] 表示 s1[0...i-1]变换到  s2[0...j-1] 的最短编辑距离 
char s1[1005], s2[1005];

inline int min(int a, int b)
{
    return a < b ? a : b;
}

inline int max(int a, int b)
{
    return a > b ? a : b;
}

void init()
{
    memset(dp, 0, sizeof(dp));
    int tmp = max(len1, len2);
    for(int i = 1; i <= tmp; ++i)  // 注意初始化 
    {
        dp[i][0] = dp[0][i] = i;
    }
}

int DP()
{
    for(int i = 0; i < len1; ++i)
        for(int j = 0; j < len2; ++j)
        {
            if(s1[i] == s2[j])
                dp[i+1][j+1] = min(min(dp[i+1][j]+1, dp[i][j+1]+1), dp[i][j]);
            else 
                dp[i+1][j+1] = min(min(dp[i+1][j]+1, dp[i][j+1]+1), dp[i][j]+1); 
        }
    return dp[len1][len2];
}

int main()
{
    while(cin >> len1 >> s1)
    {
        cin >> len2 >> s2;
        init();
        cout << DP() << endl;
    }
    return 0;
}