POJ 3356 （最长公共子序列）

Let x and y be two strings over some finite alphabet A. We would like to transform xinto 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

 【题解】 找最少操作数，使得串1等于串2，就是求出两串的最长公共子序列，长串减去公共串长度即为所求。

特别注意这题的坑点是要加入多组数据输入，不然会WA。

 【AC代码】

#include<iostream>#include<stdio.h>#include<map>#include<algorithm>#include<math.h>#include<queue>#include<stack>#include<string>#include<cstring>using namespace std;int m,n,dp[1001][1001];char g[1001],l[1001];int main(){    while(~scanf("%d",&m))    {        memset(dp,0,sizeof dp);        scanf("%s",g);        scanf("%d",&n);        scanf("%s",l);        for(int i=0;i<m;i++)        {            for(int j=0;j<n;j++)            {                if(g[i]==l[j])                    dp[i+1][j+1]=dp[i][j]+1;                else                    dp[i+1][j+1]=max(dp[i+1][j],dp[i][j+1]);            }        }        int p=max(m,n);        printf("%d\n",p-dp[m][n]);    }    return 0;}