HDU - 1560----DNA sequence
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The twenty-first century is a biology-technology developing century. We know that a gene is made of DNA. The nucleotide bases from which DNA is built are A(adenine), C(cytosine), G(guanine), and T(thymine). Finding the longest common subsequence between DNA/Protein sequences is one of the basic problems in modern computational molecular biology. But this problem is a little different. Given several DNA sequences, you are asked to make a shortest sequence from them so that each of the given sequence is the subsequence of it.
For example, given “ACGT”,”ATGC”,”CGTT” and “CAGT”, you can make a sequence in the following way. It is the shortest but may be not the only one.
Input
The first line is the test case number t. Then t test cases follow. In each case, the first line is an integer n ( 1<=n<=8 ) represents number of the DNA sequences. The following k lines contain the k sequences, one per line. Assuming that the length of any sequence is between 1 and 5.
Output
For each test case, print a line containing the length of the shortest sequence that can be made from these sequences.
Sample Input14ACGTATGCCGTTCAGTSample Output8
题目大意:给n个序列,找到一个包含所有给出序列的最短长度并输出。
思路:开始看到题目时,就直接的准备DFS最多40层,后来发现了(迭代加深搜索)其实可以减少许多无谓的搜索,注意剪枝.
#include<iostream>#include<cstring>#include<stdio.h>using namespace std; const int N=8;char str[N][5];int n,ans,depth,size[N];//限制深度char DNA[] = {'A','G','C','T'};void DFS(int tier,int len[]){ int temLen = 0;//预计还要匹配的字符串的最大长度为未匹配到的字符串长度最 for(int i=0 ; i<n ; i++){ if(size[i]-len[i] > temLen){ temLen = size[i]-len[i]; } } if(temLen == 0){//不需要匹配说明已找到 ans = tier; return; } if(tier+temLen > depth){//预计的层+已经搜的层>限制深度,停止搜索 return; } for(int i=0 ; i<4 ; i++){ int newLen[10]; bool flag = false; for(int j=0 ; j<n ; j++){ if(str[j][len[j]] == DNA[i]){//匹配到一个 flag = true; newLen[j] = len[j]+1;//更新匹配长度 }else{ newLen[j] = len[j]; } } if(flag){//没有更新匹配长度说明此情况以搜索过 DFS(tier+1,newLen); } if(ans != 0){//已找到答案,停止搜索 break; } }}int main(void){ int c; cin>>c; while(c--){ cin>>n; depth=0;//初始限制深度 for(int i=0 ; i<n ; i++){ cin>>str[i]; size[i] = strlen(str[i]); if(size[i] > depth){//初始限制深度为最长的字符串长度 depth = size[i]; } } int len[N]={0};//记录了n个字符串已经匹配到的位置 ans = 0; while(true){ DFS(0,len); if(ans == 0){ depth++;//如过没有找到答案,则搜索加深一层 }else{ cout<<ans<<endl; break; } } } return 0;}
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