poj3080 Blue Jeans【KMP】【暴力】
阿新 • • 發佈:2018-11-28
Blue Jeans
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 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:
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions:21746 | Accepted: 9653 |
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.
Input
- 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
South Central USA 2006
題意:
給定$m$個場長度為$60$的字串,問他們的最長公共子串。
思路:
因為$m$和長度都很小,所以可以暴力列舉一個串的所有子串,$KMP$進行匹配。
等之後【字尾陣列】刷熟練一點了再用字尾陣列做一次。奇怪的是hdu2328明明是一樣的題意,怎麼就總是WA
1 #include<iostream> 2 //#include<bits/stdc++.h> 3 #include<cstdio> 4 #include<cmath> 5 #include<cstdlib> 6 #include<cstring> 7 #include<algorithm> 8 #include<queue> 9 #include<vector> 10 #include<set> 11 #include<climits> 12 #include<map> 13 using namespace std; 14 typedef long long LL; 15 typedef unsigned long long ull; 16 #define pi 3.1415926535 17 #define inf 0x3f3f3f3f 18 19 const int maxn = 65; 20 int n, m; 21 char str[15][maxn]; 22 int nxt[maxn]; 23 24 void getnxt(char *s) 25 { 26 int len = strlen(s); 27 nxt[0] = -1; 28 int k = -1; 29 int j = 0; 30 while(j < len){ 31 if(k == -1 || s[j] == s[k]){ 32 ++k;++j; 33 if(s[j] != s[k]){ 34 nxt[j] = k; 35 } 36 else{ 37 nxt[j] = nxt[k]; 38 } 39 } 40 else{ 41 k = nxt[k]; 42 } 43 } 44 } 45 46 bool kmp(char *s, char *t) 47 { 48 getnxt(s); 49 int slen = strlen(s), tlen = strlen(t); 50 int i = 0, j = 0; 51 while(i < slen && j < tlen){ 52 if(j == -1 || s[i] == t[j]){ 53 j++; 54 i++; 55 } 56 else{ 57 j = nxt[j]; 58 } 59 } 60 if(j == tlen){ 61 return true; 62 } 63 else{ 64 return false; 65 } 66 } 67 68 int main() 69 { 70 scanf("%d", &n); 71 while(n--){ 72 scanf("%d", &m); 73 for(int i = 0; i < m; i++){ 74 scanf("%s", str[i]); 75 } 76 77 char ans[maxn]; 78 int anslen = -1; 79 for(int i = 0; i < 60; i++){ 80 for(int j = i; j < 60; j++){ 81 char t[maxn]; 82 memcpy(t, str[0] + i, j - i + 1); 83 t[j - i + 1] = '\0'; 84 bool flag = true; 85 for(int k = 1; k < m; k++){ 86 if(!kmp(str[k], t)){ 87 flag = false; 88 break; 89 } 90 } 91 if(flag){ 92 if(j - i + 1 > anslen){ 93 anslen = j - i + 1; 94 strcpy(ans, t); 95 } 96 else if(j - i + 1 == anslen){ 97 if(strcmp(ans, t) > 0){ 98 strcpy(ans, t); 99 } 100 } 101 } 102 else{ 103 break; 104 } 105 } 106 } 107 108 if(anslen < 3){ 109 printf("no significant commonalities\n"); 110 } 111 else{ 112 //cout<<ans<<endl; 113 printf("%s\n", ans); 114 } 115 116 } 117 return 0; 118 }