POJ 3080 字符串水题

Blue Jeans

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 15076 Accepted: 6691

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

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

题意：

求几个串的公共连续字串，如果这个公共串长度小于三就不算。如果有很多个最长长度的串。那么输出长度最短的一个。

题解：

拿到题看到60单位的字符串还是有点吃惊。想是不是KMP啊。 。trie树啊。 。dp啊什么。 。后来发现直接暴力就好。 。虽然是有O（n*k^3）的复杂度。 。但是k = 60，n = 10总共才200w+，比计算机的运算速度还是小。直接枚举子集就可以了。这里有两个地方值得注意：

一是strncpy是可拿来吧s2的一段长度复制给s1。但是这个操作是不会带上’\0’的。这点与strcpy不一样。

二是求字典序可以直接用C++里string类来，s1

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <string>
#define f(i,a,b) for(int i = a;i<=b;i++)
#define fi(i,a,b) for(int i = a;i>=b;i--)

using namespace std;

char G[15][65];
char tem[65];

int main()
{
//    freopen("data.in","r",stdin);
int N;
scanf("%d",&N);
while(N--){
int n;
scanf("%d",&n);
string s1 = "";
int Max = -0x3f3f3f3f;
f(i,1,n) scanf("%s",G[i]);
int len = strlen(G[1]);
int ok = 0;
f(i,0,len-1)
f(j,3,len-i){
//                memset(tem,0,sizeof(tem));
strncpy(tem,G[1]+i,j);
tem[j] = '\0';
int k;
for(k = 2;k<=n;k++)
if(!strstr(G[k],tem))
break;
if(k<=n);
else{
ok = 1;
if(j>Max){
Max = j;
s1 = tem;
}else if(j == Max){
string s2 = tem;
if(s2<s1)
s1 = s2;
}
}

}
if(ok)
cout << s1 << endl;
else
printf("no significant commonalities\n");
}
return 0;
}