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poj 3356

2024-10-12 05:05:38   211人阅读

Description

Let x and y be two strings over some finite alphabet A. We would like to transformx 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 inx is m and the number of letters in y is n wheren ≥ 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 stringx 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 stringx into a string y.

Sample Input

10 AGTCTGACGC
11 AGTAAGTAGGC

Sample Output

4
题意：
求由字符串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
变换的字符s2所须要 的最小操作次数。
思路：这是一个求编辑最短距离问题。利用动态规划，列出状态方程，设dp[i][j]表示字符串x[1...i]和字符串y[1...j]的最短编辑距离当x[i] == y[j]时，i和j不须要编辑,要么删除，要么插入。要么替换dp[i][j] = min(dp[i-1][j-1], dp[i-1][j] + 1, dp[i][j - 1] + 1)当x[i] != y[i]时, i和j不须要编辑dp[i][j] = min(dp[i-1][j-1] + 1, dp[i-1][j] + 1, dp[i][j-1] + 1);注意初始化dp[i][0] = dp[0][i] = i;
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 1010;
char strx[maxn], stry[maxn];
int lenx, leny, dp[maxn][maxn];
int main()
{

    while( scanf("%d %s", &lenx, strx  + 1) != EOF)
    {
        scanf("%d %s", &leny, stry + 1);
        int maxv = max(lenx, leny);
        dp[0][0] = 0;
        for(int i = 1; i <= maxv; i++)
            dp[0][i] = dp[i][0] = i;
        for(int i = 1; i <= lenx; i++)
        {
            for(int j = 1; j <= leny; j++)
            {
                dp[i][j] = min(dp[i-1][j] + 1, dp[i][j-1] + 1);
                if(strx[i] == stry[j])
                    dp[i][j] = min(dp[i][j], dp[i-1][j-1]);
                else
                    dp[i][j] = min(dp[i][j], dp[i-1][j-1] + 1);
            }
        }
        printf("%d\n", dp[lenx][leny]);
    }

    return 0;
}



poj 3356


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