zoukankan      html  css  js  c++  java
  • POJ3080-Blue Jeans

    题目链接:点击打开链接

    Blue Jeans

    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 20430   Accepted: 9053

    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
    
    

    题目大意:找3个所给字符串的最长公共子序列,(长度大于等于3)

    思路:拿第一个字符串,暴力取字串,长度递增,去其他字符串里面找。

    AC代码:

    #include<iostream>
    #include<string>
    #include<cstdio>
    #include<vector> 
    #include<algorithm>
    using namespace std;
    
    string str[15];
    string temp;
    vector<string> t;
    vector<string> ans;
    
    bool cmp(string a, string b) {//优先顺序
    	if(a.size() == b.size())
    		return a < b;
    	else
    		return a.size() > b.size();
    }
    
     
    
    int main() {
    	int n, m;
    	while(cin >> n) {
    		while(n--) {
    			cin >> m;
    			for(int i = 0; i < m; i++) {//存储
    				cin >> str[i];
    			}
    			ans.clear();
    			for(int j = 3; j <= 60; j++) {//最短3,最长60
    				for(int i = 0; i <= str[0].size() - j; i++) {//注意等号
    					int flag = 0;
    					temp = str[0].substr(i, j);//从i开始,取长度为j的字串
    					for(int k = 1; k < m; k++) {//去其他字符串里面找
    						if(str[k].find(temp) == string::npos) {
    							flag = 1;
    							break;
    						}
    					}
    					if(flag == 0)
    						ans.push_back(temp);//合法的
    				}
    			}
    			sort(ans.begin(), ans.end(), cmp);//第一个就是
    			if(ans.size() != 0)
    				if(ans[0].size() >= 3)
    					cout << ans[0] << endl;
    				else
    					cout << "no significant commonalities" << endl;
    			else
    				cout << "no significant commonalities" << endl;
    		}
    	}
    }
    
  • 相关阅读:
    ESXi创建磁盘命令
    TNS-12518,TNS-12536,TNS-00506,Linux Error: 11: Resource temporarily unavailable
    监听的instance status blocked分析
    Oracle 用户、对象权限、系统权限
    MIME详解
    11g等待事件之library cache: mutex X
    Latch Free
    PowerDesigner小技巧
    yum本地源配置
    内核参数SEMMSL SEMMNS SEMOPM SEMMNI参数的设置
  • 原文地址:https://www.cnblogs.com/ACMerszl/p/9572974.html
Copyright © 2011-2022 走看看