HDU 6138 AC自動機

題意：

題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=6138
給出n個字符串，m個詢問，詢問x和y兩個字符的最長公共子串的長度，而且要求這個子串爲這n個字符串中任意一個（或多個）的前綴。

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思路：

AC自動機。
每個節點end數組都保存當前節點到根節點的距離。將n個字符串保加入AC自動機。然後對x在自動機上跑一邊標記節點，然後再對y跑一遍，遇到標記的節點就更新答案。注意每次還要再去除標記。

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代碼：

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int INF = 0x3f3f3f3f;
const LL MOD = 1e9 + 7;
const int MAXN = 1e5 + 10;

struct ACauto {
    int next[MAXN][26], fail[MAXN], end[MAXN], vis[MAXN];
    int root, sz;

    int newnode() {
        for (int i = 0; i < 26; i++)
            next[sz][i] = -1;
        end[sz++] = 0;
        return sz - 1;
    }

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

    int idx(char c) {
        return c - 'a';
    }

    void insert(char *buf) {
        int len = strlen(buf);
        int now = root;
        for (int i = 0; i < len; i++) {
            int id = idx(buf[i]);
            if (next[now][id] == -1)
                next[now][id] = newnode();
            now = next[now][id];
            end[now] = i + 1;
            vis[now] = false;
        }
    }

    void build() {
        queue <int> Q;
        fail[root] = root;
        for (int i = 0; i < 26; 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 < 26; 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 query(char *s, int ls, char *t, int lt) {
        int now = root, res = 0;
        for (int i = 0; i < ls; i++) {
            int id = idx(s[i]);
            now = next[now][id];
            int tmp = now;
            while (tmp != root) {
                vis[tmp] = true;
                tmp = fail[tmp];
            }
        }
        now = root;
        for (int i = 0; i < lt; i++) {
            int id = idx(t[i]);
            now = next[now][id];
            int tmp = now;
            while (tmp != root) {
                if (vis[tmp]) res = max(res, end[tmp]);
                tmp = fail[tmp];
            }
        }
        now = root;
        for (int i = 0; i < ls; i++) {
            int id = idx(s[i]);
            now = next[now][id];
            int tmp = now;
            while (tmp != root) {
                vis[tmp] = false;
                tmp = fail[tmp];
            }
        }
        return res;
    }

} ac;

string s[MAXN];
int len[MAXN];
char s1[MAXN], s2[MAXN];

int main() {
    //freopen("in.txt", "r", stdin);
    int T;
    scanf("%d", &T);
    while (T--) {
        int n, m;
        scanf("%d", &n);
        ac.init();
        for (int i = 1; i <= n; i++) {
            cin >> s[i];
            len[i] = s[i].length();
            strcpy(s1, s[i].c_str());
            ac.insert(s1);
        }
        ac.build();
        /*for (int i = 0; i < ac.sz; i++)
            cout << i << " " << ac.end[i] << endl;*/
        scanf("%d", &m);
        while (m--) {
            int x, y;
            scanf("%d%d", &x, &y);
            strcpy(s1, s[x].c_str());
            strcpy(s2, s[y].c_str());
            printf("%d\n", ac.query(s1, len[x], s2, len[y]));
        }
    }
    return 0;
}