Blue Jeans
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 4119 | Accepted: 1707 |
Description
The Genographic Project is a research partnership between IBM and The National Geographic Society that is analyzing DNA from hundreds of thousands of contributors to map how the Earth was populated.
As an IBM researcher, you have been tasked with writing a program that will find commonalities amongst given snippets of DNA that can be correlated with individual survey information to identify new genetic markers.
A DNA base sequence is noted by listing the nitrogen bases in the order in which they are found in the molecule. There are four bases: adenine (A), thymine (T), guanine (G), and cytosine (C). A 6-base DNA sequence could be represented as TAGACC.
Given a set of DNA base sequences, determine the longest series of bases that occurs in all of the sequences.
As an IBM researcher, you have been tasked with writing a program that will find commonalities amongst given snippets of DNA that can be correlated with individual survey information to identify new genetic markers.
A DNA base sequence is noted by listing the nitrogen bases in the order in which they are found in the molecule. There are four bases: adenine (A), thymine (T), guanine (G), and cytosine (C). A 6-base DNA sequence could be represented as TAGACC.
Given a set of DNA base sequences, determine the longest series of bases that occurs in all of the sequences.
Input
Input to this problem will begin with a line containing a single integer n indicating the number of datasets. Each dataset consists of the following components:
- A single positive integer m (2 <= m <= 10) indicating the number of base sequences in this dataset.
- m lines each containing a single base sequence consisting of 60 bases.
Output
For each dataset in the input, output the longest base subsequence common to all of the given base sequences. If the longest common subsequence is less than three bases in length, display the string "no significant commonalities" instead. If multiple subsequences of the same longest length exist, output only the subsequence that comes first in alphabetical order.
Sample Input
3 2 GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA 3 GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA GATACTAGATACTAGATACTAGATACTAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA GATACCAGATACCAGATACCAGATACCAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA 3 CATCATCATCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC ACATCATCATAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AACATCATCATTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT
Sample Output
no significant commonalities AGATAC CATCATCAT
Source
首先把字符串连接起来，求sa,rank.height
二分答案，然后对height进行分组，在同一个分组中统计是否满足条件。
1 #include <iostream>
2 #include <cstring>
3 #include <cstdio>
4
5 using namespace std;
6 #define Max 610
7 int mem[4][Max];
8 int *rank;
9 int *sa;
10 int *nrank;
11 int *nsa;
12 int height[Max];
13 int cnt[Max];
14 int n;
15
16 char s[Max];
17 void make_sa()
18 {
19 memset(mem,0,sizeof(mem));
20 sa = mem[0];
21 rank = mem[1];
22 nsa = mem[2];
23 nrank = mem[3];
24 memset(cnt, 0, sizeof(cnt));
25 for (int i = 0; i < n; ++i)
26 {
27 ++cnt[(int)s[i]];
28 }
29 for (int i = 1; i < 256; ++i)
30 {
31 cnt[i] += cnt[i-1];
32 }
33 for (int i = n-1; i >= 0; --i)
34 {
35 sa[--cnt[(int)s[i]]] = i;
36 }
37 rank[sa[0]] = 0;
38 for (int i = 1; i < n; ++i)
39 {
40 rank[sa[i]] = rank[sa[i-1]];
41 if (s[sa[i]] != s[sa[i-1]])
42 {
43 ++rank[sa[i]];
44 }
45 }
46 for (int k = 1; k < n && rank[sa[n-1]] < n-1; k *= 2)
47 {
48 for (int i = 0; i < n; ++i)
49 {
50 cnt[rank[sa[i]]] = i + 1;
51 }
52 for (int i = n-1; i >= 0; --i)
53 {
54 if (sa[i]-k >= 0)
55 {
56 nsa[--cnt[rank[sa[i]-k]]] = sa[i]-k;
57 }
58 }
59 for (int i = n-k; i < n; ++i)
60 {
61 nsa[--cnt[rank[i]]] = i;
62 }
63 nrank[nsa[0]] = 0;
64 for (int i = 1; i < n; ++i)
65 {
66 nrank[nsa[i]] = nrank[nsa[i-1]];
67 if (rank[nsa[i]] != rank[nsa[i-1]] || rank[nsa[i]+k] != rank[nsa[i-1]+k])
68 {
69 ++nrank[nsa[i]];
70 }
71 }
72 swap(rank, nrank);
73 swap(sa, nsa);
74 }
75 }
76
77 void get_lcp_rmq()
78 {
79 int i, j, k;
80 for (i = 0, k = 0; i < n; ++i)
81 {
82 if (rank[i]==0)
83 {
84 height[0] = k = 0;
85 }
86 else
87 {
88 if (k > 0)
89 {
90 --k;
91 }
92 j = sa[rank[i]-1];
93 for (; s[i+k] == s[j+k]; ++k);
94 height[rank[i]] = k;
95 }
96 }
97 }
98 int num;
99 int stMark;
100 bool solve(int len)
101 {
102 // printf("ss\n");
103 for(int i=1;i<n;i++)
104 {
105 bool flag[10];
106 memset(flag,false,sizeof(flag));
107 int t;
108 while(i<n&&height[i]>=len)
109 {
110 flag[sa[i-1]/60] = true;
111 flag[sa[i]/60] = true;
112 t = sa[i-1];
113 i++;
114 }
115 int tcnt = 0;
116 for(int i=0;i<10;i++)
117 {
118 if(flag[i])
119 {
120 tcnt++;
121 }
122 }
123 if(tcnt==num)
124 {
125 stMark = t;
126 return true;
127 }
128
129 }
130 return false;
131 }
132 void work()
133 {
134 int l = 3,r = 61;
135 bool mark = false;
136 int ans=0;
137 while(l<r)
138 {
139 int m = l+(r-l)/2;
140 if(solve(m)){ans=m;mark = true;l = m+1;}
141 else r = m;
142 }
143 if(!mark)
144 {
145 printf("no significant commonalities\n");
146 }
147 else
148 {
149 int st = stMark;
150 int cnt = 0;
151 while(cnt<ans)
152 {
153 printf("%c",s[st]);
154 cnt++;
155 st++;
156 }
157 printf("\n");
158 }
159 }
160 int main()
161 {
162 int T;
163 scanf("%d",&T);
164 while(T--)
165 {
166 memset(s,0,sizeof(s));
167 memset(height,0,sizeof(height));
168 scanf("%d",&num);
169 int st = 0;
170 for(int i=0;i<num;i++)
171 {
172 scanf("%s",s+st);
173 st+=60;
174 s[st++] = i;
175 }
176 n = st;
177 make_sa();
178 get_lcp_rmq();
179 work();
180 }
181 return 0;
182 }
183
2 #include <cstring>
3 #include <cstdio>
4
5 using namespace std;
6 #define Max 610
7 int mem[4][Max];
8 int *rank;
9 int *sa;
10 int *nrank;
11 int *nsa;
12 int height[Max];
13 int cnt[Max];
14 int n;
15
16 char s[Max];
17 void make_sa()
18 {
19 memset(mem,0,sizeof(mem));
20 sa = mem[0];
21 rank = mem[1];
22 nsa = mem[2];
23 nrank = mem[3];
24 memset(cnt, 0, sizeof(cnt));
25 for (int i = 0; i < n; ++i)
26 {
27 ++cnt[(int)s[i]];
28 }
29 for (int i = 1; i < 256; ++i)
30 {
31 cnt[i] += cnt[i-1];
32 }
33 for (int i = n-1; i >= 0; --i)
34 {
35 sa[--cnt[(int)s[i]]] = i;
36 }
37 rank[sa[0]] = 0;
38 for (int i = 1; i < n; ++i)
39 {
40 rank[sa[i]] = rank[sa[i-1]];
41 if (s[sa[i]] != s[sa[i-1]])
42 {
43 ++rank[sa[i]];
44 }
45 }
46 for (int k = 1; k < n && rank[sa[n-1]] < n-1; k *= 2)
47 {
48 for (int i = 0; i < n; ++i)
49 {
50 cnt[rank[sa[i]]] = i + 1;
51 }
52 for (int i = n-1; i >= 0; --i)
53 {
54 if (sa[i]-k >= 0)
55 {
56 nsa[--cnt[rank[sa[i]-k]]] = sa[i]-k;
57 }
58 }
59 for (int i = n-k; i < n; ++i)
60 {
61 nsa[--cnt[rank[i]]] = i;
62 }
63 nrank[nsa[0]] = 0;
64 for (int i = 1; i < n; ++i)
65 {
66 nrank[nsa[i]] = nrank[nsa[i-1]];
67 if (rank[nsa[i]] != rank[nsa[i-1]] || rank[nsa[i]+k] != rank[nsa[i-1]+k])
68 {
69 ++nrank[nsa[i]];
70 }
71 }
72 swap(rank, nrank);
73 swap(sa, nsa);
74 }
75 }
76
77 void get_lcp_rmq()
78 {
79 int i, j, k;
80 for (i = 0, k = 0; i < n; ++i)
81 {
82 if (rank[i]==0)
83 {
84 height[0] = k = 0;
85 }
86 else
87 {
88 if (k > 0)
89 {
90 --k;
91 }
92 j = sa[rank[i]-1];
93 for (; s[i+k] == s[j+k]; ++k);
94 height[rank[i]] = k;
95 }
96 }
97 }
98 int num;
99 int stMark;
100 bool solve(int len)
101 {
102 // printf("ss\n");
103 for(int i=1;i<n;i++)
104 {
105 bool flag[10];
106 memset(flag,false,sizeof(flag));
107 int t;
108 while(i<n&&height[i]>=len)
109 {
110 flag[sa[i-1]/60] = true;
111 flag[sa[i]/60] = true;
112 t = sa[i-1];
113 i++;
114 }
115 int tcnt = 0;
116 for(int i=0;i<10;i++)
117 {
118 if(flag[i])
119 {
120 tcnt++;
121 }
122 }
123 if(tcnt==num)
124 {
125 stMark = t;
126 return true;
127 }
128
129 }
130 return false;
131 }
132 void work()
133 {
134 int l = 3,r = 61;
135 bool mark = false;
136 int ans=0;
137 while(l<r)
138 {
139 int m = l+(r-l)/2;
140 if(solve(m)){ans=m;mark = true;l = m+1;}
141 else r = m;
142 }
143 if(!mark)
144 {
145 printf("no significant commonalities\n");
146 }
147 else
148 {
149 int st = stMark;
150 int cnt = 0;
151 while(cnt<ans)
152 {
153 printf("%c",s[st]);
154 cnt++;
155 st++;
156 }
157 printf("\n");
158 }
159 }
160 int main()
161 {
162 int T;
163 scanf("%d",&T);
164 while(T--)
165 {
166 memset(s,0,sizeof(s));
167 memset(height,0,sizeof(height));
168 scanf("%d",&num);
169 int st = 0;
170 for(int i=0;i<num;i++)
171 {
172 scanf("%s",s+st);
173 st+=60;
174 s[st++] = i;
175 }
176 n = st;
177 make_sa();
178 get_lcp_rmq();
179 work();
180 }
181 return 0;
182 }
183
