poj3080Blue Jeans
題目鏈接：
啊哈哈，點我點我
題意是：
給n個字符串然後找出n個字符串裡面最長的公共字串。。
這道題目最開始以為是dp，後來又以為是kmp，但是kmp貌似沒看到過這麼多字符串相匹配的，後來就搜題解，太弱了，只能看別人題解。。
思路是：
首先看數據大小，最多只有10個串，那麼把第一個串當作母串，然後逐個去枚舉母串中的子串，然後根據字串去其他n-1個DNA序列中檢測，看是否這些子串在其他DNA序列中存在，然後把第一個母串中所有的的字串進行枚舉，得到所有字符串都滿足的最長公共字串，如果存在相同的子串，那麼選取字典序最小的那個。。這樣這題就已暴力的方式得到了解決。。然後就是找枚舉的字串在其他DNA序列中不需要像BF算法那樣一個個好，然後回溯，因為cstring頭文件裡面有一個神器 strstr。。這回真是開了眼界了。。。
題目：
Blue Jeans
Time Limit: 1000MS
Memory Limit: 65536K
Total Submissions: 12149
Accepted: 5266
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
代碼為：
#include
#include
#include
#include
using namespace std;
const int len=60;
char DNA[10+10][len+1];
char ans[len+1],Copy[len+1];
int ans_length,length;
int main()
{
int t,n,pd,flag,i,j,k,count;
scanf("%d",&t);
while(t--)
{
count=0;
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%s",DNA[i]);
ans_length=-1;
length=1;
for(i=0;;i++)
{
flag=1;
pd=i;
if(pd+length>len)//判斷序列是否越界
{
length++;
i=-1;
if(length>len)
break;
continue;
}
for(j=0;j0)
strcpy(ans,Copy);
}
if(length>ans_length)
{
ans_length=length;
strcpy(ans,Copy);
}
}
}
if(ans_length<3)
printf("no significant commonalities\n");
else
printf("%s\n",ans);
}
return 0;
}
