poj 3080 Blue Jeans 直接枚举

Language: Default

Blue Jeans

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12472		Accepted: 5409

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

Source

South Central USA 2006

题意：给出m个长度为60的DNA序列，要求出他们的最长的公共子串，若有多个长度相同的就输出字典序最小的一个。这一题我开始还在用KMP匹配，老是Wa，也不知道哪里写错了，后来想到不是有strstr函数吗，我真是二了。。。

代码有点乱，可能不是很好看。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 1005
#define MAXN 2005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
typedef long long ll;
using namespace std;

int n,m;
char str[65];
char ps[65];
char pstr[12][65];
char ans[65];

int main()
{
    int i,j;
    scanf("%d",&n);
    while (n--)
    {
        scanf("%d",&m);
        scanf("%s",str);
        for (i=0;i<m-1;i++)
            scanf("%s",pstr[i]);
        int ok=0;
        for (i=60;i>=3;i--)   //从大到小枚举长度，遇到符合要求的就输出，否则接着往下找
        {
            ans[0]='Z';ans[1]='\0';
            for (j=0;j<=60-i;j++)
            {
                for (int t=j;t<i+j;t++)  //取出长度为i的子串
                    ps[t-j]=str[t];
                ps[i]='\0';
                int flag=1;
                for (int t=0;t<m-1;t++)  //将取出来的子串与其他DNA序列比较，看是否包含子串
                {
                    if (!strstr(pstr[t],ps))
                    {
                        flag=0;
                        break;
                    }
                }
                if (flag)
                {
                    ok=1;
                    if (strcmp(ans,ps)>0)  //ans数组记录下字典序最小的子串
                        strcpy(ans,ps);
                }
            }
            if (ans[0]!='Z')
            {
                printf("%s\n",ans);
                break;
            }
        }
        if (!ok)
            printf("no significant commonalities\n");
    }
    return 0;
}
/*
4
2
GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
3
GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA
GATACTAGATACTAGATACTAGATACTAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA
GATACCAGATACCAGATACCAGATACCAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA
3
CATCATCATCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC
ACATCATCATAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AACATCATCATTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT
2
GATGATGATGATGATGATGATGATGATGATGATGATGATGATGATGATGATGATGATGAT
GATTGATTGATTGATTGATTGATTGATTGATTGATTGATTGATTGATTGATTGATTGATT
*/