HDU 3247 Resource Archiver

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

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
const int size = 2;
const char base = '0';
const int maxn = 1005;
int n, m;

class tire
{
public:
    tire *next[size], *fail;
    int cnt, virus, deep, id;
    tire(){ cnt = -1; id = virus = 0; fail = NULL; memset(next, 0, sizeof(next)); }
};

class ac
{
private:
    tire *root, *node[15];
    int map[15][15], dp[1 << 11][11], tot,vis[60005];
    char s[maxn];
public:
    ac(){ memset(map, 0, sizeof(map));  root = new tire; tot = 0; }
    void clear(){ memset(map, 0, sizeof(map)); root = new tire; tot = 0; }
    void insert(int x)
    {
        scanf("%s", s);
        tire *j = root;
        for (int i = 0, k; s[i]; i++)
        {
            k = s[i] - base;
            if (!j->next[k])
            {
                j->next[k] = new tire;
                j->next[k]->id = ++tot;
            }
            j = j->next[k];
        }
        if (x < 0) j->virus = 1; else
        {
            node[x] = j;
            j->cnt  = x;
            map[x][x] = strlen(s);
        }
    }
    void getfail()
    {
        root->fail = root;
        queue<tire *> p;
        for (int i = 0; i < size; i++)
        {
            if (root->next[i])
            {
                root->next[i]->fail = root;
                p.push(root->next[i]);
            }
            else root->next[i] = root;
        }
        tire *q, *k;
        while (!p.empty())
        {
            q = p.front();    p.pop();
            k = q->fail;
            for (int i = 0; i < size; i++)
            if (q->next[i])
            {
                q->next[i]->fail = k->next[i];
                p.push(q->next[i]);
            }
            else q->next[i] = k->next[i];
        }
    }
    void getmap()
    {
        queue<tire *> p;
        tire *q, *k;
        for (int i = 0; i < n; i++)
        {
            p.push(node[i]);
            node[i]->deep = 0;
            memset(vis, 0, sizeof(vis));
            vis[node[i]->id] = 1;
            while (!p.empty())
            {
                q = p.front();    p.pop();
                for (int j = 0; j < size; j++)
                if (!vis[q->next[j]->id])
                {
                    k = q->next[j];
                    k->deep = q->deep + 1;
                    if (k->cnt >= 0 && !map[i][k->cnt]) map[i][k->cnt] = k->deep;
                    if (!k->virus){ p.push(k); vis[k->id] = 1; }
                }
            }
        }
    }
    int work_out()
    {
        getfail();
        getmap();
        memset(dp, 1, sizeof(dp));
        for (int i = 0; i < n; i++) dp[1 << i][i] = map[i][i];
        for (int i = 1; i < (1 << n); i++)
        {
            for (int j = 0; j < n; j++)
                if (i&(1 << j))
                    for (int k = 0; k < n;k++)
                        if ((i&(1 << k))==0)
                        if (map[j][k])
                        {
                            dp[i | (1 << k)][k] = min(dp[i | (1 << k)][k], dp[i][j] + map[j][k]);
                        }
        }
        int ans = 0x7FFFFFFF;
        for (int i = 0; i < n; i++) ans = min(ans, dp[(1 << n) - 1][i]);
        return ans;
    }
}f;

int main()
{
    while (scanf("%d%d", &n, &m), n + m)
    {
        f.clear();
        for (int i = 0; i < n; i++) f.insert(i);
        while (m--) f.insert(-1);
        printf("%d\n", f.work_out());
    }
}