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POJ 3356 AGTC(DP-最小编辑距离)

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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。那么弱要删除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):

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 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 using namespace std;
 6 
 7 const int MAXN = 1010;
 8 
 9 char s1[MAXN], s2[MAXN];
10 int dp[MAXN][MAXN];
11 int n, m;
12 
13 int main() {
14     while(scanf("%d%s", &n, s1 + 1) != EOF) {
15         scanf("%d%s", &m, s2 + 1);
16         dp[0][0] = 0;
17         for(int i = 1; i <= n; ++i) dp[i][0] = i;
18         for(int j = 1; j <= m; ++j) dp[0][j] = j;
19         for(int i = 1; i <= n; ++i) {
20             for(int j = 1; j <= m; ++j) {
21                 dp[i][j] = min(dp[i - 1][j] + 1, dp[i][j - 1] + 1);
22                 dp[i][j] = min(dp[i][j], dp[i - 1][j - 1] + (s1[i] != s2[j]));
23             }
24         }
25         printf("%d\n", dp[n][m]);
26     }
27 }
View Code

 

POJ 3356 AGTC(DP-最小编辑距离),布布扣,bubuko.com

POJ 3356 AGTC(DP-最小编辑距离)

原文:http://www.cnblogs.com/oyking/p/3698185.html

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