首页 > 代码库 > POJ 3356 AGTC（最长公共子序列）

POJ 3356 AGTC（最长公共子序列）

2024-07-16 09:04:23   222人阅读

AGTC

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

题意  给你两个DNA序列  求第一个第一个序列至少经过多次删除 、替换 或添加碱基得到第二个序列    其实分析一下可以发现   只要求出两个序列的最长公共子序列  这部分就可以不动了  然后较长序列的长度减去最长公共子序列的长度就是答案了

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 1005;
int la, lb, d[N][N];
char a[N], b[N];

void lcs()
{
    memset (d, 0, sizeof (d));
    for (int i = 1; i <= la; ++i)
        for (int j = 1; j <= lb; ++j)
            if (a[i] == b[j]) d[i][j] = d[i - 1][j - 1] + 1;
            else d[i][j] = max (d[i - 1][j], d[i][j - 1]);
}

int main()
{
    while (~scanf ("%d%s%d%s", &la, a + 1, &lb, b + 1))
    {
        lcs();
        printf ("%d\n", max (la, lb) - d[la][lb]);
    }
    return 0;
}



联系
我们