HDU 1560 sequence

题目链接
Problem Description

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 Input

1
4
ACGT
ATGC
CGTT
CAGT

Sample Output

大致题意
输入t，接下来 t 组，输入一个n，表示 n 个DNA，DNA只包含ACGT四个字母，求一个最短序列顺序包含这 n 个字母

具体思路
一开始以为是一道大模拟的题目，WA了好多发，最后发现是深搜减枝，一开始我们求出所有给定的序列最大长度，先假设这些序列所需要的最大顺序序列是maxdeep（所有序列中的最大长度），然后枚举这个情况下的排列ACGT，当出现形成之后无法包含所有的序列，此时maxdeep+1,进行下一轮的深搜枚举，直到在maxdeep的情况下可以顺序包含所有的序列后输出，一味的深搜必然会引起超时，在这里我们就需要剪枝，当深搜的时候，每当枚举一个元素，原序列中可能会有序列的未匹配首字母与其对应，对应之后会剩下还有几个字母，当这些序列的maxnun（最大剩余字母量）加上当前deep（已经匹配的最短序列）之和，大于maxdeep时，则表示我们此时的maxdeep已经无法满足需求，直接减枝进行下一轮的深搜，而当我们maxnum（最大剩余字母量）等于0的时候，则表示已经全部匹配，可以输出d=maxdeep了

#include<bits/stdc++.h>
using namespace std;
char dna[4]= {'A','C','G','T'};
int nowpot[10];
int t,maxdeep,maxlen,n;
struct C
{char import[10];int pot;
} c[10];
int maxnum()
{int sum=0;for (int i=0;i<n;i++){sum=max(sum,c[i].pot-nowpot[i]);}return sum;
}
int dfs(int deep)
{if (deep+maxnum()>maxdeep){return 0;}if (maxnum()==0){return 1;}int now[10];for (int i=0; i<n; i++){now[i]=nowpot[i];}for (int i=0; i<4; i++){int flag=0;for (int j=0; j<n; j++){if (c[j].import[nowpot[j]]==dna[i]){flag=1;nowpot[j]++;}}if (flag==1){if (dfs(deep+1)==1){return 1;}for (int k=0; k<n; k++){nowpot[k]=now[k];}}}return 0;
}
int main()
{scanf("%d",&t);while (t--){maxdeep=maxlen=0;scanf("%d",&n);for (int i=0; i<n; i++){scanf("%s%*c",&c[i].import);c[i].pot=strlen(c[i].import);maxlen=max(maxlen,c[i].pot);nowpot[i]=0;}maxdeep=maxlen;while (true){if (dfs(0)==1){break;}maxdeep++;}printf("%d\n",maxdeep);}return 0;
}

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