SPOJ - PHRASES K - Relevant Phrases of Annihilation

K - Relevant Phrases of Annihilation

题目大意：给你 n 个串，问你最长的在每个字符串中出现两次且不重叠的子串的长度。

思路：二分长度，然后将height分块，看是否存在一个块里面 每个串都符合条件。

#include<bits/stdc++.h>

#define LL long long

#define fi first

#define se second

#define mk make_pair

#define pii pair<int, int>

#define y1 skldjfskldjg

#define y2 skldfjsklejg

using namespace std;

const int N = 1e5 + ;

const int M = 1e5 + ;

const int inf = 0x3f3f3f3f;

const LL INF = 0x3f3f3f3f3f3f3f3f;

char s[N], str[N];

int n, m, b[N];

int sa[N], t[N], t2[N], c[N], rk[N], height[N];

int L[], R[];

void buildSa(char *s, int n, int m) {

    int i, j = , k = , *x = t, *y = t2;

    for(i = ; i < m; i++) c[i] = ;

    for(i = ; i < n; i++) c[x[i] = s[i]]++;

    for(i = ; i < m; i++) c[i] += c[i - ];

    for(i = n - ; i >= ; i--) sa[--c[x[i]]] = i;

    for(int k = ; k <= n; k <<= ) {

        int p = ;

        for(i = n - k; i < n; i++) y[p++] = i;

        for(i = ; i < n; i++) if(sa[i] >= k) y[p++] = sa[i] - k;

        for(i = ; i < m; i++) c[i] = ;

        for(i = ; i < n; i++) c[x[y[i]]]++;

        for(i = ; i < m; i++) c[i] += c[i - ];

        for(i = n - ; i >= ; i--) sa[--c[x[y[i]]]] = y[i];

        swap(x, y);

        p = ; x[sa[]] = ;

        for(int i = ; i < n; i++) {

            if(y[sa[i - ]] == y[sa[i]] && y[sa[i - ] + k] == y[sa[i] + k])

                x[sa[i]] = p - ;

            else x[sa[i]] = p++;

        }

        if(p >= n) break;

        m = p;

     }

     for(i = ; i < n; i++) rk[sa[i]] = i;

     for(i = ; i < n - ; i++) {

        if(k) k--;

        j = sa[rk[i] - ];

        while(s[i + k] == s[j + k]) k++;

        height[rk[i]] = k;

     }

}

bool check(int len, int n) {

    int l = , r;

    while(l <= n) {

        for(int i = ; i <= m; i++) L[i] = inf, R[i] = -inf;

        r = l;

        L[b[sa[l]]] = min(L[b[sa[l]]], sa[l]);

        R[b[sa[l]]] = max(R[b[sa[l]]], sa[l]);

        while(r < n && height[r + ] >= len) {

            r++;

            L[b[sa[r]]] = min(L[b[sa[r]]], sa[r]);

            R[b[sa[r]]] = max(R[b[sa[r]]], sa[r]);

        }

        bool flag = true;

        for(int i = ; i <= m; i++) {

            if(L[i] + len > R[i]) {

                flag = false;

                break;

            }

        }

        if(flag) return true;

        l = r + ;

    }

    return false;

}

int main() {

    int T; scanf("%d", &T);

    while(T--) {

        scanf("%d", &m);

        char ch = 'A'; int tot = ;

        for(int i = ; i <= m; i++) {

            scanf("%s", str);

            int len = strlen(str);

            for(int j = ; j < len; j++) {

                s[tot] = str[j];

                b[tot++] = i;

            }

            s[tot++] = ch++;

        }

        s[tot] = '\0';

        buildSa(s, tot + , );

        int l = , r = tot, mid, ans = ;

        while(l <= r) {

            mid = l + r >> ;

            if(check(mid, tot)) ans = mid, l = mid + ;

            else r = mid - ;

        }

        printf("%d\n", ans);

    }

    return ;

}

/*

1

4

abbabba

dabddkababa

bacaba

baba

*/