HDU 6138 Fleet of the Eternal Throne （ AC自动机）

HDU 6138 Fleet of the Eternal Throne （ AC自动机）。

Problem Description > The Eternal Fleet was built many centuries ago before the time of Valkorion by an unknown race on the planet of Iokath. The fate of the Fleet's builders is unknown but their legacy would live on. Its first known action was in the annihilation of all life in Wild Space. It spread across Wild Space and conquered almost every inhabited world within the region, including Zakuul. They were finally defeated by a mysterious vessel known as the Gravestone, a massive alien warship that countered the Eternal Fleet's might. Outfitted with specialized weapons designed to take out multiple targets at once, the Gravestone destroyed whole sections of the fleet with a single shot. The Eternal Fleet was finally defeated over Zakuul, where it was deactivated and hidden away. The Gravestone landed in the swamps of Zakuul, where the crew scuttled it and hid it away.
>
> — Wookieepedia

The major defeat of the Eternal Fleet is the connected defensive network. Though being effective in defensing a large fleet, it finally led to a chain-reaction and was destroyed by the Gravestone. Therefore, when the next generation of Eternal Fleet is built, you are asked to check the risk of the chain reaction.
The battleships of the Eternal Fleet are placed on a 2D plane ofnrows. Each row is an array of battleships. The type of a battleship is denoted by an English lowercase alphabet. In other words, each row can be treated as a string. Below lists a possible configuration of the Eternal Fleet.
aa
bbbaaa
abbaababa
abba


If in thex-th row and they-th row, there exists a consecutive segment of battleships that looks identical in both rows (i.e., a common substring of thex-th row andy-th row), at the same time the substring is a prefix of any other row (can be thex-th or they-th row), the Eternal Fleet will have a risk of causing chain reaction.

Given a query (x,y), you should find the longest substring that have a risk of causing chain reaction.
Input The first line of the input contains an integerT, denoting the number of test cases.

For each test cases, the first line contains integern(n≤105).

There arenlines following, each has a string consisting of lower case letters denoting the battleships in the row. The total length of the strings will not exceed105.

And an integerm(1≤m≤100) is following, representing the number of queries.

For each of the followingmlines, there are two integersx,y, denoting the query.
Output You should output the answers for the queries, one integer per line.
Sample Input

1 3 aaa baaa caaa 2 2 3 1 2

Sample Output

3 3

Source 2017 Multi-University Training Contest - Team 8
Recommend liuyiding|We have carefully selected several similar problems for you:61436142614161406139

题解：对n个串造出 自动AC机 ，这里的AC机不需要维护单词的结束点 需要维护一个每个节点到根的距离，也就是前缀的长度。然后用x跑一遍AC机，在所有匹配成功的结束节点上记上一个标记（直接记成询问次数就行了：第一次询问flag记成1，第二次记成2，这样保证每次的flag都不一样就不用清空标记了），然后这些标记点的意思就是：这个前缀是n个串中某个的前缀，且这个前缀是串x的字串。然后再用y跑一次AC机，在y匹配成功的节点，如果他刚刚被打了标记，那就统计最长的length就可以。对于每个节点都要打标记，不是在字母最后一个地方打。。

原来模板太不好变化了。。有改了下

#include 
#include 
#include 
#include 
#include 
using namespace std;
const int maxn = 1e6 + 7;
const int maxnode = 50*10010;
int n;
char s1[60], s2[maxn];
int pos[maxn];
struct node
{
    int ch[maxnode][26], cnt[maxnode], fail[maxnode], last[maxnode], id, road[maxnode]; // 路径压缩优化， 针对模式串出线的种类
    int dep[maxnode], flag[maxnode];
    void init()
    {
        memset(ch[0], 0, sizeof(ch[0]));
        memset(dep, 0, sizeof(dep));
        id = 1;
    }
    void Insert(char *s)
    {
        int rt = 0;
        int len = strlen(s);
        dep[rt] = 0;
        for(int i = 0; i < len; i++)
        {
            if(!ch[rt][s[i]-'a'])
            {
                memset(ch[id], 0, sizeof(ch[id]));
                flag[id] = 0;
                ch[rt][s[i]-'a'] = id++;
            }
            dep[ch[rt][s[i]-'a']] = dep[rt] + 1;
            rt = ch[rt][s[i]-'a'];
        }
    }
    void get_fail()
    {
        queue q;
        fail[0] = 0;
        int rt = 0;
        for(int i = 0; i < 26; i++)
        {
            rt = ch[0][i];
            if(rt)
            {
                q.push(rt);
                fail[rt] = 0;
            }
        }
        while(!q.empty())
        {
            int r = q.front();
            q.pop();
            for(int i = 0; i < 26; i++)
            {
                rt = ch[r][i];
                if(!rt)
                {
                    ch[r][i] = ch[fail[r]][i];
                    continue;
                }
                q.push(rt);
                fail[ch[r][i]] = ch[fail[r]][i];
            }
        }
    }
    void Match(char *s, int f)
    {
        int rt = 0;
        int len = strlen(s);
        for(int i = 0; i < len; i++)
        {
            int temp = rt = ch[rt][s[i]-'a'];
            while(temp)
            {
                flag[temp] = f;
                temp = fail[temp];
            }
        }
    }
    int Search(char *s, int f)
    {
        int rt = 0, res = 0;
        int len = strlen(s);
        for(int i = 0; i < len; i++)
        {
            int temp = rt = ch[rt][s[i]-'a'];
            while(temp)
            {
                if(flag[temp] == f) res = max(res, dep[temp]);
                temp = fail[temp];
            }
        }
        return res;
    }
}ac_auto;
char s[maxn];
int main()
{
    int t, q;
    cin >> t;
    while(t--)
    {
        scanf("%d", &n);
        int d = 0;
        ac_auto.init();
        for(int i = 0; i < n; i++)
        {
            pos[i] = d;
            scanf(" %s", s+d);
            ac_auto.Insert(s+d);
            int len = strlen(s+d);
            d += len + 1;
        }
        ac_auto.get_fail();
        scanf("%d", &q);
        int ca = 1;
        while(q--)
        {
            int x, y;
            scanf("%d%d", &x, &y);
            x--; y--;
            ac_auto.Match(s+pos[x],ca);
            int ans = ac_auto.Search(s+pos[y], ca++);
            printf("%d\n", ans);
        }
    }
    return 0;
}