POJ 3356 AGTC （最长公共子序列）

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

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]); } }