POJ 1699 二進位制表示狀態+dfs

好長時間沒有做計劃了啊，這道題目還是放假之前看的，一放假亂七八糟的啊，沒有dbug，今天看了一下，手殘了一點啊。

題意很好理解：就是給你n串字母看怎麼才能做成最小的一串字元。期間用到了二進位制表示剩餘的所有的子狀態：http://blog.csdn.net/fulongxu/article/details/22953717。然後再dfs列舉就可以了啊。注意如果不能合併的就不用再算了啊。

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

Output

Sample Input

1
5
TCGG
GCAG
CCGC
GATC
ATCG

Sample Output

11

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-7
#define M 1000100
#define LL __int64
//#define LL long long
#define INF 0x3fffffff
#define PI 3.1415926535898

using namespace std;

const int maxn = 30;

char str[maxn][maxn];
int dis[maxn][maxn];
int vis[maxn];
int f[2010];
int cnt;
int Min;

int judge(char s1[], char s2[])
{
    int len1 = strlen(s1);
    int len2 = strlen(s2);
    int ans = 0;
    int i, j;
    for(i = 0; i < len1; i++)
    {
        if(s1[i] == s2[0])
        {
            for(j = 1; j < len2 && i+j < len1; j++)
            {
                if(s1[i+j] != s2[j])
                    break;
            }
            if(j == len2 || i+j == len1)
                break;
        }
    }
    ans = len2-(len1-i);
    return ans;
}


void dfs(int x, int len, int s)
{
    s = s-(1<<x);
    int flag = 0;
    int ts = s;
    while(s)
    {
        int tt = (s-1)&s;
        int ss = s-tt;
        int xx = f[ss];
        s = s - (1<<xx);
        dfs(xx, len+dis[x][xx], ts);
        flag = 1;
    }
    if(!flag)
        Min = min(Min, len);
}


int main()
{
    for(int i = 0; i <= 10; i++)
        f[(1<<i)] = i;

    for(int i = 1; i <= (1<<10); i++)
        if(!f[i]) f[i] = f[i-1];
    int T;
    cin >>T;
    while(T--)
    {
        int n;
        cin >>n;
        for(int i = 0; i < n; i++)
            cin >>str[i];
        Min = INF;
        memset(vis, 0 , sizeof(vis));
        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < n; j++)
            {
                if(i == j || vis[i] || vis[j])
                {
                    continue;
                }
                else
                {
                    int ans = judge(str[i], str[j]);
                    dis[i][j] = ans;
                    if(ans <= 0)
                        vis[j] = 1;
                }
            }
        }

        int ss = (1<<n)-1;
        for(int i = 0; i < n; i++)
            if(vis[i]) ss -= (1<<i);
        for(int i = 0; i < n; i++)
            if(!vis[i]) dfs(i, strlen(str[i]), ss);
        cout<<Min<<endl;
    }
}