Resource Archiver

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 100000/100000 K (Java/Others)
Total Submission(s): 2382    Accepted Submission(s): 750

Problem Description

Great! Your new software is almost finished! The only thing left to do is archiving all your n resource files into a big one.
Wait a minute… you realized that it isn’t as easy as you thought. Think about the virus killers. They’ll find your software suspicious, if your software contains one of the m predefined virus codes. You absolutely don’t want this to happen.
Technically, resource files and virus codes are merely 01 strings. You’ve already convinced yourself that none of the resource strings contain a virus code, but if you make the archive arbitrarily, virus codes can still be found somewhere.
Here comes your task (formally): design a 01 string that contains all your resources (their occurrences can overlap), but none of the virus codes. To make your software smaller in size, the string should be as short as possible.

Input

There will be at most 10 test cases, each begins with two integers in a single line: n and m (2 <= n <= 10, 1 <= m <= 1000). The next n lines contain the resources, one in each line. The next m lines contain the virus codes, one in each line. The resources and virus codes are all non-empty 01 strings without spaces inside. Each resource is at most 1000 characters long. The total length of all virus codes is at most 50000. The input ends with n = m = 0.

Output

For each test case, print the length of shortest string.

Sample Input

2 2

1110

0111

101

1001

0 0

Sample Output

5

/*
hdu 3247 AC自动+状压dp+bfs处理

给你n个正常子串,m个病毒子串,求出最短的字符串(包含所有正常子串,不包含病毒串)

因为正常子串只有十个,所以考虑二进制来记录。
即dp[i][j]表示 包含的正常串的状态为i,最后一步的状态为j的最短情况.
然后试了下发现超内存 卒~

然后膜拜大神,发现我们可以预处理出来正常串之间的最短距离. 像这样我们只需要枚举所有的
正常串. 而我原先那个思路需要枚举所有的节点总共需要dp[1<<10][60040].  而对于通过bfs优化后
我们只需要枚举正常串 最多有11个 ->  dp[1<<10][11].

hhh-2016-04-30 14:34:51
*/
#include <iostream>
#include <vector>
#include <cstring>
#include <string>
#include <cstdio>
#include <queue>
#include <algorithm>
#include <functional>
#include <map>
using namespace std;
#define lson  (i<<1)
#define rson  ((i<<1)|1)
typedef unsigned long long ll;
typedef unsigned int ul;
const int mod = 20090717;
const int INF = 0x3f3f3f3f;
const int N = 100050;
int cnt;
int n,m;
int dp[1<<10][205];
int G[205][205];
int pos[205];
struct Tire
{
    int nex[N][2],fail[N],ed[N];

    int root,L;
    int newnode()
    {
        for(int i = 0; i < 2; i++)
            nex[L][i] = -1;
        ed[L++] = 0;
        return L-1;
    }

    void ini()
    {
        L = 0,root = newnode();
    }

    void inser(char buf[],int val)
    {
        int len = strlen(buf);
        int now = root;
        for(int i = 0; i < len; i++)
        {
            int ta = buf[i]-‘0‘;
            if(nex[now][ta] == -1)
                nex[now][ta] = newnode();
            now = nex[now][ta];
        }
        if(val < 0)
        ed[now] = val;
        else
            ed[now] = (1<<val);
    }

    void build()
    {
        queue<int >q;
        fail[root] = root;
        for(int i = 0; i < 2; i++)
            if(nex[root][i] == -1)
                nex[root][i] = root;
            else
            {
                fail[nex[root][i]] = root;
                q.push(nex[root][i]);
            }
        while(!q.empty())
        {
            int now = q.front();
            q.pop();
            if(ed[fail[now]] < 0)
                ed[now] = ed[fail[now]];
            else if(ed[now] == 0)
                ed[now] = ed[fail[now]];
            for(int i = 0; i < 2; i++)
            {
                if(nex[now][i] == -1)
                    nex[now][i] = nex[fail[now]][i];
                else
                {
                    fail[nex[now][i]] = nex[fail[now]][i];
                    q.push(nex[now][i]);
                }
            }
        }
    }
    int dis[N];
    void Path(int k)
    {
        int now;
        queue<int >q;
        q.push(pos[k]);
        memset(dis,-1,sizeof(dis));
        dis[pos[k]] = 0;
        while(!q.empty())
        {
            now = q.front();
            q.pop();
            for(int i =0;i < 2;i++)
            {
                int t = nex[now][i];
                if(dis[t] < 0 && ed[t] >= 0)
                {
                    dis[t] = dis[now]+1;
                    q.push(t);
                }
            }
        }
        for(int i = 0;i < cnt;i++)
        {
            G[k][i] = dis[pos[i]];
        }
    }

    int Min(int a,int b)
    {
        if(a < 0)
            return b;
        else if(b < 0)
            return a;
        else
            return min(a,b);
    }

    void solve()
    {
        memset(dp,-1,sizeof(dp));
        dp[0][0] = 0;

        for(int i = 0;i < (1<<n);i++)
        {
            for(int j = 0;j < cnt;j++)
            {
                if(dp[i][j] < 0)
                    continue;
                for(int k = 0;k < cnt;k++)
                {
                    if(G[j][k] < 0)
                        continue;
                    int t = (i|ed[pos[k]]);
                    dp[t][k] = Min(dp[i][j] + G[j][k],dp[t][k]);
                }
            }
        }
        int ans = -1;
        for(int i = 0;i < cnt;i++)
        {
            ans = Min(ans,dp[(1<<n)-1][i]);
        }
        cout << ans <<"\n";
    }
};

Tire ac;
char buf[N];
int main()
{
    while(scanf("%d%d",&n,&m)==2 && n && m)
    {
        ac.ini();
        for(int i = 0; i < n; i++)
        {
            scanf("%s",buf);
            ac.inser(buf,i);
        }
        for(int i =0 ; i < m; i++)
        {
            scanf("%s",buf);
            ac.inser(buf,-1);
        }
        ac.build();
        pos[0] = 0;
        cnt = 1;
        for(int i = 0; i < ac.L; i++)
        {
            if(ac.ed[i] > 0)
                pos[cnt++] = i;
        }
        memset(G,-1,sizeof(G));
        for(int i = 0; i < cnt; i++)
            ac.Path(i);
        ac.solve();
    }
    return 0;
}