POJ 3356 AGTC （最长公共子序列）

朴灿烈づ我的快乐病毒、 2024-02-17 19:42 1阅读 0赞

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

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

思路：最长公共子序列

AC代码：

#include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    const int maxn=1000+10;
    char x[maxn],y[maxn];
    int dp[maxn][maxn];
    int main(){
    	int a1,b1;
    	while(scanf("%d",&a1)==1){
    		scanf("%s",x+1);
    		scanf("%d%s",&b1,y+1);
    		
    		memset(dp,0,sizeof(dp));
    		for(int i=1;i<=a1;i++){
    			for(int j=1;j<=b1;j++){
    				if(x[i]==y[j])
    				dp[i][j]=dp[i-1][j-1]+1;
    				else {
    					dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
    				}
    			}
    		}
    		
    		int ans=max(a1,b1);
    		printf("%d\n",ans-dp[a1][b1]);
    	}
    }