Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 12744 | Accepted: 4801 |
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.
IllustrationA 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 | | | | | | | 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
题意:给两个字符串,计算由第一个变成第二个的最小操作次数,只可以删除,添加和转换操作。思路:略。
# include <iostream> # include <cstdio> # include <cstring> # include <algorithm> # define MAXN 1000 using namespace std; char a[MAXN+3], b[MAXN+3]; int dp[MAXN+3][MAXN+3]; int main() { int n, m; while(~scanf("%d%s%d%s",&n,a+1, &m, b+1)) { memset(dp, 0, sizeof(dp)); for(int i=1; i<=m; ++i) dp[0][i] = i; for(int i=1; i<=n; ++i) dp[i][0] = i; for(int i=1; i<=n; ++i) for(int j=1; j<=m; ++j) { dp[i][j] = dp[i-1][j-1] + (a[i]!=b[j]);//转换 dp[i][j] = min(min(dp[i][j], dp[i-1][j]+1), dp[i][j-1]+1);//删除和增加 } printf("%d\n",dp[n][m]); } return 0; }