POJ 3356 AGTC 解題報告（Edit Distance，DP）

AGTC Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 9579 Accepted: 3659 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 AGTCTGACGC 11 AGTAAGTAGGC Sample Output 4 Source Manila 2006

    解題報告： 今天比賽被坑的一個地方，題目中出現了Edit Distance，卻不知道具體是個什麼東西，一直以為用LCS就可以搞定，然後一直WA。 

    以為是LCS也有先入為主的思想在裡面，聽到别人這麼說，就直接以為就是這樣，沒有去想别的。

    Edit Distance就是兩個串通過添加一個字元，删除一個字元，修改一個字元這樣的變換，得到相同的串的步驟數。

    直接DP就好了。代碼如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

char a[1111], b[1111];
int dp[1111][1111];

int dfs(int l, int r)
{
    if(l==0 && r==0) return 0;
    if(l==0 && r) return r;
    if(l && r==0) return l;

    if(~dp[l][r]) return dp[l][r];
    int &ans=dp[l][r];

    ans=min(min(dfs(l-1, r)+1, dfs(l, r-1)+1), dfs(l-1, r-1)+(a[l]!=b[r]));
    return ans;
}

int main()
{
    int lena, lenb;
    while(~scanf("%d%s%d%s", &lena, a+1, &lenb, b+1))
    {
        memset(dp, -1, sizeof(dp));
        printf("%d\n", dfs(lena, lenb));
    }
}