POJ3080 Blue Jeans —— 暴力枚举 + KMP / strstr()

题目链接：https://vjudge.net/problem/POJ-3080

 

Blue Jeans

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 19152		Accepted: 8524

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

 

 

题解：

找出所有字符串的最长公共连续子串。

直接枚举第一个字符串的每个子串，然后通过kmp算法或者strstr()函数，取判断该子串是否存在于每一个字符串。

 

KMP：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <sstream>
#include <algorithm>
using namespace std;
typedef long long LL;
const double eps = 1e-6;
const int INF = 2e9;
const LL LNF = 9e18;
const int MOD = 1e9+7;
const int MAXN = 100+10;

char s[MAXN][MAXN], tmp[MAXN], ans[MAXN];
int Next[MAXN], Len[MAXN];

void get_next(char x[], int m)
{
    int i, j;
    j = Next[0] = -1;
    i = 0;
    while(i<m)
    {
        while(j!=-1 && x[i]!=x[j]) j = Next[j];
        Next[++i] = ++j;
    }
}

bool kmp(char x[], int m, char y[], int n)
{
    int i, j;
    get_next(x, m);
    i = j = 0;
    while(i<n)
    {
        while(j!=-1 && y[i]!=x[j]) j = Next[j];
        i++; j++;
        if(j>=m) return true;
    }
    return false;
}

int main()
{
    int T, n;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d", &n);
        for(int i = 1; i<=n; i++)
            scanf("%s", s[i]), Len[i] = strlen(s[i]);

        bool hav_ans;
        for(int len = Len[1]; len>=2; len--)
        {
            hav_ans = false;
            for(int st = 0; st<=Len[1]-len; st++)
            {
                int en = st+len-1, cnt = 0;
                for(int i = st; i<=en; i++)
                    tmp[cnt++] = s[1][i];
                tmp[cnt] = 0;

                bool all = true;
                for(int i = 2; i<=n; i++)
                    all = all&&kmp(tmp, cnt, s[i], Len[i]);

                if(all)
                {
                    if(!hav_ans || strcmp(tmp, ans)<0)
                    {
                        strcpy(ans, tmp);
                        hav_ans = true;
                    }
                }
            }
            if(hav_ans) break;
        }
        if(hav_ans) puts(ans);
        else puts("no significant commonalities");
    }
}

 

strstr()：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <sstream>
#include <algorithm>
using namespace std;
typedef long long LL;
const double eps = 1e-6;
const int INF = 2e9;
const LL LNF = 9e18;
const int MOD = 1e9+7;
const int MAXN = 100+10;

char s[MAXN][MAXN], tmp[MAXN], ans[MAXN];

int main()
{
    int T, n;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d", &n);
        for(int i = 1; i<=n; i++)
            scanf("%s", s[i]);

        int Len = strlen(s[1]);
        bool hav_ans;
        for(int len = Len; len>=2; len--)
        {
            hav_ans = false;
            for(int st = 0; st<=Len-len; st++)
            {
                int en = st+len-1, cnt = 0;
                for(int i = st; i<=en; i++)
                    tmp[cnt++] = s[1][i];
                tmp[cnt] = 0;

                bool all = true;
                for(int i = 2; i<=n; i++)
                    all = all&&(strstr(s[i], tmp));

                if(all)
                {
                    if(!hav_ans || strcmp(tmp, ans)<0)
                    {
                        strcpy(ans, tmp);
                        hav_ans = true;
                    }
                }
            }
            if(hav_ans) break;
        }
        if(hav_ans) puts(ans);
        else puts("no significant commonalities");
    }
}