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 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 nwhere 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： 把一个字符串经过最少操作步数转为另一个字符串 ——------操作可以是删除插入修改一个字符
dp[i][j]表示A[0-i] B[0-j]相等的最少步数
我们先来对B进行操作
删除的B[j] : d[i][j]=d[i][j-1]+1;
在B[j]后面插入一个： d[i][j]=d[i-1][j]+1;
删除一个数 if(B[j]==A[i]) d[i][j]=d[i-1][j-1];
else d[i][j]=d[i-1][j-1]+1;
求以上三种方法的最大d[i][j];
同理对A[i]操作也方程也是不变的
#include<iostream>
#include<cstring>
#include<string>
using namespace std;
int dp[][];
string a,b;
int n,m;
int Dp(int i,int j){
if(dp[i][j]==-){
int t1=Dp(i-,j)+;
int t2=Dp(i,j-)+;
int t3=Dp(i-,j-)+(a[i-]==b[j-]?:);
dp[i][j]=min(min(t1,t2),t3);
}
return dp[i][j];
}
int main(){
while(cin>>n>>a>>m>>b){
memset(dp,-,sizeof(dp));
int t=max(n,m);
for(int i=;i<=t;i++){
dp[][i]=i;
dp[i][]=i;
}
cout<<Dp(n,m)<<endl;
}
return ;
}