Best Sequence_DFS&&KMp

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

1
5
TCGG
GCAG
CCGC
GATC
ATCG

Sample Output

11

【题意】找出最短的字符串长度，要求包含给出的所有字符串

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
const int inf=0x7777777;
int n,maxn;
int mp[][],ln[],vis[],next1[][];
char s[][];
int kmp(int x,int y)
{
    int i=,j=;
    while(i<ln[x])
    {
        if(s[x][i]==s[y][j])
        {
            i++;j++;
        }
        else
        {
            if(j==) i++;
            else
            {
                j=next1[x][j];
            }
        }
        if(j==ln[y]) return -;//能在x中找到y；
    }
    return j;
}
void get_next(int k)
{
    int j=-;
    next1[k][]=-;
    for(int i=;i<ln[k];i++)
    {
        if(j==-||s[k][i]==s[k][j]) next1[k][++i]=++j;
        else j=next1[k][j];
    }
}
void dfs(int len,int cnt,int k)
{
    if(cnt==n)//已经是第n个字符串了，获取最小的长度
    {
        if(len<maxn) maxn=len;
        return;
    }
    for(int i=;i<=n;i++)
    {
        if(vis[i]==)
        {
            vis[i]=;
            if(mp[k][i]==-)//如果字符串k和i包含关系
            {
                dfs(len,cnt+,k);//最后一个参数仍然是k
            }
            else dfs(len+ln[i]-mp[k][i],cnt+,i);//如果不包含，两个串的长度相加减去相同部分长度
            vis[i]=;
        }
    }
    return;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=;i<=n;i++)
        {
            scanf("%s",s[i]);
            ln[i]=strlen(s[i]);
            get_next(i);//求next数组
        }
        for(int i=;i<=n;i++)
        {
            for(int j=;j<=n;j++)
            {
                mp[i][j]=kmp(i,j);//求两个字符串的匹配度
            }
        }
        maxn=inf;
        memset(vis,,sizeof(vis));
        for(int i=;i<=n;i++)
        {
            vis[i]=;
            dfs(ln[i],,i);
            vis[i]=;
        }
        printf("%d\n",maxn);
    }
    return ;
}