zoukankan      html  css  js  c++  java
  • poj

    poj - 3080-Blue Jeans

                                                Blue Jeans

    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 17434   Accepted: 7728

    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

    16558378 sunshinestyle 3080 Accepted 700K 16MS G++ 1245B 2017-02-08 22:06:52

    题解:

    初级题目, 采用暴力即可。求出两两的最长连续子串,逐个匹配。 

    //p-3080  
    
    #include <iostream> 
    #include <cstdio> 
    #include <cstdlib> 
    #include <cstring> 
    using namespace std; 
    const int MAXN = 65; 
    const int MAXM = 11; 
    
    int dp[MAXN][MAXN];  
    char data[MAXM][MAXN], LCS[MAXN], comp[MAXN]; 
    
    int function(int idx){
    	memset(dp, 0, sizeof(dp));
    	int max_len = 0, pt = -1;   
    	for(int i=1; i<=60; ++i){
    		for(int j=1; j<=strlen(comp+1); ++j){
    			if(data[idx][i] == comp[j]){
    				dp[i][j] = dp[i-1][j-1] + 1; 
    				if(dp[i][j] > max_len){
    					max_len = dp[i][j]; 
    					pt = j; 
    				} else if(dp[i][j] == max_len){
    					if(comp[j-max_len+1] < comp[pt-max_len+1]){
    						pt = j; 
    					}
    				}
    			}
    		}
    	}
    	if(max_len >= 3){
    		for(int i=1; i<=max_len; ++i){
    			LCS[i] = comp[pt-max_len+i]; 
    		}
    		LCS[max_len+1] = ''; 
    	}
    	return max_len; 
    }
    
    int main(){
    	freopen("in.txt", "r", stdin); 
    
    	int test_num, m; 
    	scanf("%d", &test_num); 
    	while(test_num--){
    		scanf("%d", &m); 
    		scanf("%s", comp+1); 
    		for(int i=0; i<m-1; ++i){
    			scanf("%s", data[i]+1); 
    		}
    		int idx = 0, flag = 1; 
    		while(idx<(m-1)){
    			if(function(idx) < 3){
    				flag = 0; 
    				break; 
    			}
    			strcpy(comp+1, LCS+1); 
    			idx++; 
    		}
    		if(flag == 0 || strlen(LCS+1) < 3){
    			printf("no significant commonalities
    ");
    		}else{
    			printf("%s
    ", LCS+1 );
    		}
    	}
    	return 0; 
    }
    

      

  • 相关阅读:
    Hive-0.12.0 配置Mysql的Metasotre
    cdecl、stdcall、fastcall、thiscall函数调用约定区别 (转)
    汇编函数 哪些寄存器在使用时需要保护和恢复现场
    如何用C++ 写Python模块扩展(二)
    如何用C++ 写Python模块扩展(一)
    python装饰器装饰原理探秘
    Linux 命令
    iOS为所需要的视图添加模糊效果--UIVisualEffectView
    UIAlertView ----警告视图
    Virtual Box 下Ubuntu桥接网络设置
  • 原文地址:https://www.cnblogs.com/zhang-yd/p/6380085.html
Copyright © 2011-2022 走看看