(简单dp) poj 3356 AGTC

AGTC

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5058 Accepted: 1953

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

/*题目大意：求串S变成T的最少改变次数.(变换，删除，插入)。题目解答：Dp简单题Source CodeProblem: 3356  User: wawadimu Memory: 4136K  Time: 16MS Language: C++  Result: Accepted Source Code */#include<iostream>using namespace std;#define maxn 1010char a[maxn],b[maxn];int ans[maxn][maxn];int min(int x,int y) {return x>y ? y:x;}int main(){//freopen("3356.txt","r",stdin);int n,m;int i,j,k;while(scanf("%d ",&n)!=EOF){for(i=0;i<n;i++){//cin>>a[i];scanf(" %c",a+i);//cout<<a[i];}//cout<<a[n-1]<<endl;scanf("/n%d ",&m);for(i=0;i<m;i++){//cin>>b[i];scanf("%c",b+i);}for(i=0;i<=n;i++)//注意下标ans[i][0]=i;for(j=0;j<=m;j++)ans[0][j]=j;for(i=1;i<=n;i++){for(j=1;j<=m;j++){if(a[i-1]==b[j-1])ans[i][j]=min(min(ans[i-1][j-1],ans[i-1][j]+1),ans[i][j-1]+1);elseans[i][j]=min(min(ans[i-1][j-1]+1,ans[i-1][j]+1),ans[i][j-1]+1);}}printf("%d/n",ans[n][m]);}return 0;}