POJ_3356——最短编辑距离,动态规划

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


关于最短编辑距离可以看一下我的蓝桥杯子栏中的文章，那里有些分析
这题我只是用来测试蓝桥杯里面的“DNA比对”这题，顺便水过的 - -

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 10000;
int main()
{
    int n1, n2;
    char a[N], b[N];
    while(cin >> n1 >> a >> n2 >> b)
    {
        int d1[N + 1] = {0}, d2[N + 1] = {0};
        for(int i = 0; i <= N; i++)
        {
            d1[i] = i;
        }

        for(int i = 1; i <= n1; i++)      //a(0, i)
        {
            d2[0] = i;
            for(int j = 1; j <= n2; j++)  //b(0, j)
            {
                d2[j] = min(d2[j-1] + 1, d1[j] + 1);
                if(a[i - 1] == b[j - 1])
                {
                    d2[j] = min(d2[j], d1[j-1]);
                }
                else
                {
                    d2[j] = min(d2[j], d1[j-1] + 1);
                }
            }
            for(int k = 0; k <= N; k++)
            {
                d1[k] = d2[k];
            }
        }
        cout << d2[n2] << endl;
    }
    return 0;
}