POJ 3356 AGTC（最長公用子序列）

標籤：acm   poj   dna   字串   

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 AGTCTGACGC11 AGTAAGTAGGC

Sample Output

4

題意  給你兩個DNA序列  求第一個第一個序列至少經過多次刪除 、替換 或添加堿基得到第二個序列    其實分析一下可以發現   只要求出兩個序列的最長公用子序列  這部分就可以不動了  然後較長序列的長度減去最長公用子序列的長度就是答案了

#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;}