PKU3356 字符串 DP
AGTC
| Time Limit: 1000MS |
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Memory Limit: 65536K |
| Total Submissions: 4930 |
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Accepted: 1904 |
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 .
Sample Input
10 AGTCTGACGC
11 AGTAAGTAGGC
Sample Output
4
设有长度 l1 和 l2 的字符串 s1 和 s2,res[i][j] 为 s1 的前 i 个字符变成 s2 的前 j 个字符所需的最少操作数 , 转移方程为 :
if (s1[i]==s2[j]) res[i+1][j+1]=min(res[i][j],res[i+1][j]+1,res[i][j+1]+1);
else res[i+1][j+1]=min(res[i][j]+1,res[i+1][j]+1,res[i][j+1]+1);
当 (s1[i]==s2[j]) 时 ,res[i+1][j+1] 可以直接取 res[i][j], 也可以从 res[i+1][j] 加 1 得到 , 也可以从 res[i][j+1] 得到 , 取其中的最小者便是当前的最优解 .
(s1[i]!=s2[j]) 时相似 .
代码如下 :
#include<stdio.h> #include<string.h> #define MAXN 1005 int min(int a,int b,int c) { if (a>b) return b>c?c:b; return a>c?c:a; } int res[MAXN][MAXN]; int main() { int l1,l2,i,j; char s1[MAXN],s2[MAXN]; while (scanf("%d%s",&l1,s1)!=EOF) { scanf("%d%s",&l2,s2); for (i=0;i<=l1;++i) res[i][0]=i; for (i=0;i<=l2;++i) res[0][i]=i; for (i=0;i<l1;++i) for (j=0;j<l2;++j) if (s1[i]==s2[j]) res[i+1][j+1]=min(res[i][j],res[i+1][j]+1,res[i][j+1]+1); else res[i+1][j+1]=min(res[i][j]+1,res[i+1][j]+1,res[i][j+1]+1); printf("%d/n",res[l1][l2]); } return 0; }