POJ 1699 kmp+枚举

Best Sequence

Description

The twenty-first century is a biology-technology developing century. One of the most attractive and challenging tasks is on the gene project, especially on gene sorting program. Recently we know that a gene is made of DNA. The nucleotide bases from which DNA is built are A(adenine), C(cytosine), G(guanine), and T(thymine). Given several segments of a gene, you are asked to make a shortest sequence from them. The sequence should use all the segments, and you cannot flip any of the segments.

For example, given ‘TCGG‘, ‘GCAG‘, ‘CCGC‘, ‘GATC‘ and ‘ATCG‘, you
can slide the segments in the following way and get a sequence of length
11. It is the shortest sequence (but may be not the only one).

Input

The
first line is an integer T (1 <= T <= 20), which shows the number
of the cases. Then T test cases follow. The first line of every test
case contains an integer N (1 <= N <= 10), which represents the
number of segments. The following N lines express N segments,
respectively. Assuming that the length of any segment is between 1 and
20.

Output

For each test case, print a line containing the length of the shortest sequence that can be made from these segments.

Sample Input

1
5
TCGG
GCAG
CCGC
GATC
ATCG

Sample Output

11

Source

POJ Monthly--2004.07.18

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cstdlib>
 5 #include<algorithm>
 6 using namespace std;
 7
 8 const int inf=0x3fffffff;
 9 const int maxn=11;
10 char str[maxn][21];
11 int cost[maxn][maxn];
12 int len[maxn];
13 int n,t;
14
15 int cal(char *st1, char *st2)
16 {
17     int len1 = strlen(st1);
18     int len2 = strlen(st2);
19     for (int i = len1 - len2; i < len1; i++)
20     {
21         bool ok = true;
22         for (int j = i; j < len1; j++)
23             if (st1[j] != st2[j - i])
24             {
25                 ok = false;
26                 break;
27             }
28         if (ok)
29             return len2 + i - len1;
30     }
31     return len2;
32 }
33 void work()
34 {
35     for(int i=0;i<n;i++)
36         for(int j=0;j<n;j++)
37     {
38         cost[i][j]=cal(str[i],str[j]);
39     }
40 }
41 int Get_ans()
42 {
43     int next[maxn];int ans=inf,sum;
44     for(int i=0;i<n;i++) next[i]=i;
45     do{
46         sum=len[next[0]];
47         for(int i=0;i<n-1;i++)
48         {
49             sum+=cost[next[i]][next[i+1]];
50         }
51          ans=min(ans,sum);
52         }while(next_permutation(next,next+n));
53         return ans;
54 }
55 int main()
56 {
57     //freopen("in.txt","r",stdin);
58     scanf("%d",&t);
59     while(t--){
60             memset(str,0,sizeof(str));
61             scanf("%d",&n);
62             for(int i=0;i<n;i++)
63             {
64                 scanf("%s",str[i]);
65                 len[i]=strlen(str[i]);
66             }
67             work();
68             int ans=Get_ans();
69             printf("%d\n",ans);
70     }
71     return 0;
72 }