poj_3080Blue Jeans && poj_3450Corporate Identity(KMP)
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 9476 | Accepted: 3983 |
Description
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
Sample Input
3 2 GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA 3 GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA GATACTAGATACTAGATACTAGATACTAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA GATACCAGATACCAGATACCAGATACCAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA 3 CATCATCATCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC ACATCATCATAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AACATCATCATTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT
Sample Output
no significant commonalities AGATAC CATCATCAT枚举子串，然后KMP。
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#pragma warning(disable : 4996)
int n, m;
int Next[1000];
char str[15][65];
char ans[65];
void get_next(char *s, int m)
{
char pat[65] = {0};
strcpy(pat + 1, s);
Next[1] = 0;
int i, j = 0;
for(i = 2; i <= m; i++)
{
while(j > 0 && pat[j+1] != pat[i])
{
j = Next[j];
}
if(pat[j+1] == pat[i])
{
j += 1;
}
Next[i] = j;
}
}
bool kmp(char *A, char *B, int n, int m)
{
char pat[65] = {0};
char text[65] = {0};
strcpy(text + 1, A);
strcpy(pat + 1, B);
int i, j = 0;
for(i = 1; i <= n; i++)
{
while(j > 0 && pat[j+1] != text[i])
{
j = Next[j];
}
if(pat[j+1] == text[i])
{
j += 1;
}
if(j == m)
{
return true;
}
}
return false;
}
bool check(char *s, int tot) //匹配串tmp
{
int n, m;
m = strlen(s);
for (int i = 2; i <= tot; i++)
{
n = strlen(str[i] + 1);
if(!kmp(str[i] + 1, s, n, m))
{
return false;
}
}
return true;
}
int main()
{
freopen("in.txt", "r", stdin);
int t, n, length, len;
char tmp[65] = {0};
scanf("%d", &t);
while (t--)
{
scanf("%d", &n);
for(int i = 1; i <= n; i++)
{
scanf("%s", str[i] + 1);
}
len = strlen(str[1] + 1);
length = 0;
for(int i = 1; i <= len; i++)
{
for(int j = 1; j <= len - i + 1; j++)
{
strncpy(tmp, str[1] + j, i);
memset(Next, 0, sizeof(Next));
get_next(tmp, i);
if(check(tmp, n))
{
if(i >= length)
{
if(strcmp(tmp, ans) < 0 && i == length)
{
strcpy(ans, tmp);
length = strlen(ans);
}
else
{
strcpy(ans, tmp);
length = strlen(ans);
}
}
}
memset(tmp, 0, sizeof(tmp));
}
}
if(length >= 3)
{
printf("%s\n", ans);
}
else
{
printf("no significant commonalities\n");
}
}
return 0;
}
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 3998 | Accepted: 1514 |
Description
Beside other services, ACM helps companies to clearly state their “corporate identity”, which includes company logo but also other signs, like trademarks. One of such companies is Internet Building Masters (IBM), which has recently asked ACM for a help with their new identity. IBM do not want to change their existing logos and trademarks completely, because their customers are used to the old ones. Therefore, ACM will only change existing trademarks instead of creating new ones.
After several other proposals, it was decided to take all existing trademarks and find the longest common sequence of letters that is contained in all of them. This sequence will be graphically emphasized to form a new logo. Then, the old trademarks may still be used while showing the new identity.
Your task is to find such a sequence.
Input
The input contains several tasks. Each task begins with a line containing a positive integer N, the number of trademarks (2 ≤ N ≤ 4000). The number is followed by N lines, each containing one trademark. Trademarks will be composed only from lowercase letters, the length of each trademark will be at least 1 and at most 200 characters.
After the last trademark, the next task begins. The last task is followed by a line containing zero.
Output
For each task, output a single line containing the longest string contained as a substring in all trademarks. If there are several strings of the same length, print the one that is lexicographically smallest. If there is no such non-empty string, output the words “IDENTITY LOST” instead.
Sample Input
3 aabbaabb abbababb bbbbbabb 2 xyz abc 0
Sample Output
abb IDENTITY LOST
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#pragma warning(disable : 4996)
char str[4005][205];
char tmp[205], ans[205];
int Next[4005];
void get_next(int m, char *B)
{
char pat[205];
strcpy(pat + 1, B);
Next[1] = 0;
int i, j = 0;
for(i = 2; i <= m; i++)
{
while(j > 0 && pat[j+1] != pat[i])
{
j = Next[j];
}
if(pat[j+1] == pat[i])
{
j += 1;
}
Next[i] = j;
}
}
bool kmp(char *A, char *B, int n, int m) //主串 模式串 主串长度 模式串长度
{
char text[205], pat[205];
strcpy(text + 1, A);
strcpy(pat + 1, B);
int i, j = 0;
for(i = 1; i <= n; i++)
{
while(j > 0 && pat[j+1] != text[i])
{
j = Next[j];
}
if(pat[j+1] == text[i])
{
j += 1;
}
if(j == m)
{
return true;
}
}
return false;
}
bool check(char *s, int t)
{
for(int i = 2; i <= t; i++)
{
int n = strlen(str[i] + 1);
int m = strlen(s);
if(!kmp(str[i] + 1, s, n, m))
{
return false;
}
}
return true;
}
int main()
{
freopen("in.txt", "r", stdin);
int t, length;
while (scanf("%d", &t) != EOF)
{
if(t == 0)
{
break;
}
length = 0;
for(int i = 1; i <= t; i++)
{
scanf("%s", str[i] + 1);
}
int len = strlen(str[1] + 1);
for(int i = 1; i <= len; i++) // i 代表匹配串长度
{
for(int j = 1; j <= len - i + 1; j++)
{
strncpy(tmp, str[1] + j, i);
memset(Next, 0, sizeof(Next));
get_next(i, tmp);
if(check(tmp, t))
{
if(i >= length)
{
if(strcmp(tmp, ans) < 0 && i == length)
{
strcpy(ans, tmp);
}
if(i > length)
{
strcpy(ans, tmp);
}
length = strlen(ans);
}
}
memset(tmp, 0, sizeof(tmp));
}
}
if(strlen(ans) >= 1)
{
printf("%s\n", ans);
memset(ans, 0, sizeof(ans));
}
else
{
printf("IDENTITY LOST\n");
}
}
return 0;
}
