 poj3356---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 nm.

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`

1.转换 ，将str1和str2判断，如果相等，则dp=0，否则dp=1
2.删除，因为，目的串比源串小，所以删除源串一个字符，

3.添加，在目的串添加一个字符，即源串不变，但是目的串减1，和源串去匹配，即dp + 1

dp[i-1][j-1]+str1[i]==str2[j]?0:1
dp[i-1][j]+1
dp[i][j-1]+1

```#include

#include

using namespace std;char a,b;int dp;int main(){
int m,n;
cin>>m;
for(int i=0;i

{
cin>>a[i];
}
cin>>n;
for(int j=0;j

{
cin>>b[j];
}
dp=0;
for(int i=1;i<=max(m,n);i++)
{
dp[i]=i;
dp[i]=i;
}
for(int i=1;i<=m;i++)
{
for(int j=1;j<=n;j++)
{
if(a[i-1]==b[j-1])
{
dp[i][j]=min(dp[i-1][j-1],min(dp[i-1][j]+1,dp[i][j-1]+1));
}
else
{
dp[i][j]=min(dp[i-1][j-1],min(dp[i-1][j],dp[i][j-1]))+1;
}
}
}
cout<

return 0;}

```

#### 最新留言

[***.249.68.14]2022年05月22日 12时22分45秒

## 关于作者 喝酒易醉，品茶养心，人生如梦，品茶悟道，何以解忧？唯有杜康！
-- 愿君每日到此一游！