poj 1699 Best Sequence(dfs+A(n,n))

poj 1699 Best Sequence(dfs+A(n,n))

题目：​​Sequence

Description

The twenty-first century is a biology-technology developing century. One of the most attractive and challenging tasks is on the gene project, especially on gene sorting program. Recently 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). Given several segments of a gene, you are asked to make a shortest sequence from them. The sequence should use all the segments, and you cannot flip any of the segments.

For example, given 'TCGG', 'GCAG', 'CCGC', 'GATC' and 'ATCG', you can slide the segments in the following way and get a sequence of length 11. It is the shortest sequence (but may be not the only one).

Input

The first line is an integer T (1 <= T <= 20), which shows the number of the cases. Then T test cases follow. The first line of every test case contains an integer N (1 <= N <= 10), which represents the number of segments. The following N lines express N segments, respectively. Assuming that the length of any segment is between 1 and 20.

Output

For each test case, print a line containing the length of the shortest sequence that can be made from these segments.

Sample Input

15 TCGG GCAG CCGC GATC ATCG

Sample Output

11

分析：已知多个片段个体求极值问题，想到了动态规划。根据题意，两个字符串组合，如s1=“12345”，s2=“345678”新构成的字符串的长度比原来s1的长度要多3，把增加的长度存储在一个二维数组len[1][2]中。这样，对于以字符串si开头的由n个子串组合成的新的字符串的就是si+len(i,j,k,……)[

A(n-1,n-1)种组合情况]。那么怎样模拟这么多的组合情况呢？深度优先搜索。

#include #include #include #include using namespace std;const int INF=0x3f3f3f3f;int getmore(char *s1,char *s2){ int length1=strlen(s1),length2=strlen(s2),i; int m,qi; if(length1>length2)qi=length1-length2; else qi=0; for(;qians)return ; if(count==N && length>t; while(t--){ scanf("%d",&N); ans=INF; memset(length,0,sizeof(length)); memset(used,0,sizeof(used)); memset(len,0,sizeof(len)); for(int i=0;i

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