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POJ 3356 AGTC(DP-最小编辑距离)
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 CDeletion: * 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
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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):
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 }