Hong Kong Regional Contest 2016 [Kattis

/*项目名称：Kattis taboo创建者：  westwood创建时间：  2017/9/15 21:45:13CLR：  4.0.30319.42000解题方法：题意:给出n个01串,要构造一个最长的串使得这个串不包含所有出现过的串,无解输出-1.首先把所有的点扔进自动机,因为出现过的串不能在我们构造的串中出现,所以事先把某些”坏点”标记,坏点指的就是每个串的结束节点以及沿着结束节点fail指针走下去的节点. 如果剩下的图中能够沿着根走出一个环那么必然就无解了, 判环可以用一遍拓扑确定. 排除无解情况剩下就是一个DAG了, 最后按照最长路打印方案即可.*/#include<cstdio>#include<iostream>#include<algorithm>#include<cmath>#include<queue>#include<cstring>#include<stack>#include<vector>#pragma warning(disable : 4996)using namespace std;const static int MAXN = 500010;char str[200000 + 10];struct Tire{    int root, rcnt;    int next[MAXN][2], fail[MAXN], end[MAXN];    int newnode()    {        for (int i = 0; i < 2; i++)            next[rcnt][i] = -1;        end[rcnt++] = 0;        return rcnt - 1;    }    void init()    {        rcnt = 0;        root = newnode();    }    void insert(char buf[])    {        int len = strlen(buf);        int now = root;        for (int i = 0; i < len; i++)        {            if (next[now][buf[i] - '0'] == -1)                next[now][buf[i] - '0'] = newnode();            now = next[now][buf[i] - '0'];        }        end[now]++;    }    void build()    {        queue<int> que;        fail[root] = root;        for (int i = 0; i < 2; i++)            if (next[root][i] == -1)                next[root][i] = root;            else            {                fail[next[root][i]] = root;                que.push(next[root][i]);            }        while (que.size())        {            int now = que.front();            end[now] += end[fail[now]];//这个是包含字串，如果失配跳转的地方是一个字符串结束的位置，那么其实就是说，这个now串包含另一个字符串，匹配到这个位置也是非法的            que.pop();            for (int i = 0; i < 2; i++)            {                if (next[now][i] == -1)                    next[now][i] = next[fail[now]][i];                else                {                    fail[next[now][i]] = next[fail[now]][i];                    que.push(next[now][i]);                }            }        }    }}tire;//其他的是ac自动机模板int deg[MAXN];bool vis[MAXN];int dp[MAXN];void getdeg(){    for (int u = 0; u < tire.rcnt; u++)    {        if (tire.end[u]) continue;        for (int i = 0; i < 2; i++)        {            int v = tire.next[u][i];            if (tire.end[v]) continue;            deg[v]++;        }    }}int loop(){    memset(dp, 0, sizeof dp);    queue<int> que;    for (int i = 0; i < tire.rcnt; i++)        if (deg[i] == 0)        {            if (tire.end[i]) continue;            que.push(i), dp[i] = 1;        }    while (que.size())    {        int u = que.front(); que.pop();        for (int i = 0; i < 2; i++)        {            int v = tire.next[u][i];            if (tire.end[v]) continue;            else if (dp[v] < dp[u] + 1)            {                dp[v] = dp[u] + 1;            }            if (--deg[v] == 0) que.push(v);        }    }    int longest = 0,node;    for (int i = 0; i < tire.rcnt; i++)    {        if (deg[i]) return -1;        if (longest < dp[i])        {            longest = dp[i];            node = i;        }    }    return longest;}int longest;vector<int> path;//用外部vector保存路径bool findpath(int u)//寻找字典序最小的路径{    if (dp[u] == longest)    {        for (int i = 0; i < path.size(); i++)            printf("%d", path[i]);        exit(0);    }    for (int i = 0; i < 2; i++)    {        int v = tire.next[u][i];        if (dp[v] == dp[u] + 1)        {            path.push_back(i);            findpath(v);            path.pop_back();        }    }}void work(){    memset(vis, false, sizeof vis);    getdeg();    longest = loop();    if (longest == -1)    {        puts("-1");    }    else    {        findpath(0);    }}int main(){    if (fopen("in.txt", "r"))    {        freopen("in.txt", "r", stdin);        freopen("out.txt", "w", stdout);    }    int n;    cin >> n;    tire.init();    for (int i = 0; i < n; i++)    {        scanf("%s", str);        tire.insert(str);    }    tire.build();    work();    return 0;}