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POJ 3356-AGTC(区间DP)
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 10225 | Accepted: 3949 |
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
类似于编辑距离问题。。dp[i][j]代表 x串从1-i与y串从1-j已经匹配好的最少操作数。
#include <cstdio> #include <iostream> #include <algorithm> #include <cstring> #include <cctype> #include <cmath> #include <cstdlib> #include <vector> #include <queue> #include <set> #include <map> #include <list> #define ll long long using namespace std; const int INF = 0x3f3f3f3f; int dp[1025][1025]; char a[1025],b[1025]; int My_min(int x,int y,int z) { return min(min(x,y),z); } int main() { int lena,lenb; while(scanf("%d %s %d %s",&lena,a,&lenb,b)!=EOF){ for(int i=0;i<=lenb;i++) dp[0][i]=i; for(int i=0;i<=lena;i++) dp[i][0]=i; for(int i=1;i<=lena;i++) for(int j=1;j<=lenb;j++) dp[i][j]=My_min(dp[i-1][j]+1,dp[i][j-1]+1,a[i-1]==b[j-1]?dp[i-1][j-1]:dp[i-1][j-1]+1); printf("%d\n",dp[lena][lenb]); } return 0; }
POJ 3356-AGTC(区间DP)