POJ 3356 AGTC（DP-最小编辑距离）

时间：2021-11-23 15:43:06

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.

题目大意：给两个字符串s1、s2，输出最小编辑距离。

思路：DP，用dp[i][j]代表s1[1..i]和s2[1..j]的最小编辑距离。那么若删除s1[i]，那么dp[i][j] = dp[i - 1][j] + 1，若添加s2[j]，那么dp[i][j] = dp[i][j - 1] + 1。若用s2[j]替换s1[i]，那么dp[i][j] = dp[i - 1][j - 1] + 1。若s1[i] = s2[j]，不操作，则dp[i][j] = dp[i - 1][j - 1]。取最小值。初始化dp[0][0] = 0，dp[i][0] = i，dp[0][j] = j。复杂度$O(n^2)$。

PS：求LCS那个是错的。比如abd、acb，LCS是2，对应答案是1。但实际上答案是2，只能说明数据太水了。

代码（0MS）：

 #include <iostream>

 #include <cstdio>

 #include <cstring>

 #include <algorithm>

 using namespace std;

 const int MAXN = ;

 char s1[MAXN], s2[MAXN];

 int dp[MAXN][MAXN];

 int n, m;

 int main() {

     while(scanf("%d%s", &n, s1 + ) != EOF) {

         scanf("%d%s", &m, s2 + );

         dp[][] = ;

         for(int i = ; i <= n; ++i) dp[i][] = i;

         for(int j = ; j <= m; ++j) dp[][j] = j;

         for(int i = ; i <= n; ++i) {

             for(int j = ; j <= m; ++j) {

                 dp[i][j] = min(dp[i - ][j] + , dp[i][j - ] + );

                 dp[i][j] = min(dp[i][j], dp[i - ][j - ] + (s1[i] != s2[j]));

             }

         }

         printf("%d\n", dp[n][m]);

     }

 }

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