POJ3356AGTC题解动态规划ＤＰ

AGTC

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4870		Accepted: 1879

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

Source

Manila 2006

状态：

d[i][j]表示a串前i个字符和b串前j个字符的最小编辑距离

状态转移方程：

d[i][j]=min(d[i-1][j-1]+(a[i]!=b[j]),d[i-1][j]+1,d[i][j-1]+1)

边界：

d[i][0]=i;

d[0][j]=j;

代码：

#include<cstdio> #define min(a,b) ((a)<(b)?(a):(b)) int d[1005][1005],n,m,i,j; int main() { char a[1005],b[1005]; while(scanf("%d%s%d%s",&n,a+1,&m,b+1)==4) { for(i=0;i<=n;i++)d[i][0]=i; for(j=0;j<=m;j++)d[0][j]=j; for(i=1;a[i];i++) for(j=1;b[j];j++) d[i][j]=min(d[i-1][j-1]+(a[i]!=b[j]),min(d[i-1][j],d[i][j-1])+1); printf("%d/n",d[n][m]); } }