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:
Certainly, we would like to minimize the number of all possible operations.
IllustrationA G T A A G T * A G G C | | | | | | | A G T * C * T G A C G CDeletion: * 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;
}
POJ 3356 AGTC（最长公共子序列）,布布扣,bubuko.com
原文地址：http://blog.csdn.net/iooden/article/details/38425467
