2017 ACM-ICPC 亚洲区（青岛赛区）网络赛 HDU 6208 1003 The Dominator of Strings（AC自动机）

题意：给你n个字符串，问你其中是否存在一个字符串包含其他n-1个字符串。所有字符串加起来总长度<=1e5。

思路：如果有这个串，那肯定是n个串中最长的，如果最长的串只有一个，那将其他字符串插入ac自动机，匹配下看看匹配个数是不是n-1就行；如果最长的串有多个，这几个串必须相同，否则是NO，都相同的话跟之前一样，跑下ac自动机看看匹配数是不是n-1即可。

代码:

#include
using namespace std;
const int maxn = 1e5+5;
struct node
{
    int s, len;
    node() {}
    node(int ss, int ll): s(ss), len(ll) {}
}a[maxn];
int n, cnt, idx;
char str[maxn];
char sstr[maxn];

/****************************************************/
const int LetterSize = 26;
const int TrieSize = 10000*50+5;

int tot, root, fail[TrieSize], val[TrieSize], Next[TrieSize][LetterSize];
int newnode(void)
{
    memset(Next[tot], -1, sizeof(Next[tot]));
    val[tot] = 0;
    return tot++;
}

void init(void)
{
    tot = 0;
    root = newnode();
}

int getidx(char x)
{
    return x-'a';
}

void Insert(char *ss)
{
    int len = strlen(ss);
    int now = root;
    for(int i = 0; i < len; i++)
    {
        int idx = getidx(ss[i]);
        if(Next[now][idx] == -1)
            Next[now][idx] = newnode();
        now = Next[now][idx];
    }
    val[now]++; //和Trie一样，根据需要而变
}

void build(void)
{
    queue Q;
    fail[root] = root;
    for(int i = 0; i < LetterSize; i++)
    {
        if(Next[root][i] == -1)
            Next[root][i] = root;
        else
            fail[Next[root][i]] = root, Q.push(Next[root][i]);
    }
    while(!Q.empty())
    {
        int now = Q.front(); Q.pop();
        for(int i = 0; i < LetterSize; i++)
        {
            if(Next[now][i] == -1)
                Next[now][i] = Next[fail[now]][i];
            else
                fail[Next[now][i]] = Next[fail[now]][i], Q.push(Next[now][i]);
        }
    }
}

int match(char *ss)
{
    int len = strlen(ss), now = root, res = 0;
    for(int i = 0; i < len; i++)
    {
        int idx = getidx(ss[i]);
        int tmp = now = Next[now][idx];
        while(tmp)
        {
            res += val[tmp];
            val[tmp] = 0;
            tmp = fail[tmp];
        }
    }
    return res;
}

/*************************************************/

void solve()
{
    init();
    for(int i = 1; i <= n; i++)
    {
        if(i != idx)
        {
            for(int j = a[i].s; j < a[i].s+a[i].len; j++)
            {
                str[j-a[i].s] = sstr[j];
            }
            str[a[i].len] = 0;
            Insert(str);
        }
    }
    build();
    for(int i = a[idx].s; i < a[idx].s+a[idx].len; i++)
        str[i-a[idx].s] = sstr[i];
    str[a[idx].len] = 0;
    int num = match(str);
    if(num == n-1)
    {
        for(int i = a[idx].s; i < a[idx].s+a[idx].len; i++)
            printf("%c", sstr[i]);
        puts("");
    }
    else puts("No");
}

int main(void)
{
    int _;
    cin >> _;
    while(_--)
    {
        scanf("%d", &n);
        int p = 0;
        int maxlen = 0;
        for(int i = 1; i <= n; i++)
        {
            scanf(" %s", str);
            int len = strlen(str);
            maxlen = max(len, maxlen);
            for(int j = p; j < p+len; j++)
                sstr[j] = str[j-p];
            a[i] = node(p, len);
            p = p+len;
        }
        sstr[p] = 0;
//        puts(sstr);
        bool ok = 1;
        cnt = 0, idx = 0;
        for(int i = 1; i <= n; i++)
            if(a[i].len == maxlen)
            {
                cnt++;
                if(!idx)
                    idx = i;
            }
        if(cnt == 1) solve();
        else
        {
            for(int i = idx+1; i <= n; i++)
            {
                if(a[i].len == maxlen)
                {
                    for(int j = 0; j < maxlen; j++)
                    {
                        if(sstr[a[idx].s+j] != sstr[a[i].s+j])
                        {
                            ok = 0;
                            break;
                        }
                    }
                }
                if(!ok) break;
            }
            if(!ok) puts("No");
            else solve();
        }
    }
    return 0;
}