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 + 7;
const int M = 1e5 + 7;
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[11], R[11];

void buildSa(char *s, int n, int m) {
    int i, j = 0, k = 0, *x = t, *y = t2;
    for(i = 0; i < m; i++) c[i] = 0;
    for(i = 0; i < n; i++) c[x[i] = s[i]]++;
    for(i = 1; i < m; i++) c[i] += c[i - 1];
    for(i = n - 1; i >= 0; i--) sa[--c[x[i]]] = i;
    for(int k = 1; k <= n; k <<= 1) {
        int p = 0;
        for(i = n - k; i < n; i++) y[p++] = i;
        for(i = 0; i < n; i++) if(sa[i] >= k) y[p++] = sa[i] - k;
        for(i = 0; i < m; i++) c[i] = 0;
        for(i = 0; i < n; i++) c[x[y[i]]]++;
        for(i = 1; i < m; i++) c[i] += c[i - 1];
        for(i = n - 1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];
        swap(x, y);
        p = 1; x[sa[0]] = 0;
        for(int i = 1; i < n; i++) {
            if(y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + k] == y[sa[i] + k])
                x[sa[i]] = p - 1;
            else x[sa[i]] = p++;
        }
        if(p >= n) break;
        m = p;
     }

     for(i = 1; i < n; i++) rk[sa[i]] = i;
     for(i = 0; i < n - 1; i++) {
        if(k) k--;
        j = sa[rk[i] - 1];
        while(s[i + k] == s[j + k]) k++;
        height[rk[i]] = k;
     }
}

bool check(int len, int n) {
    int l = 1, r;
    while(l <= n) {
        for(int i = 1; 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 + 1] >= 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 = 1; i <= m; i++) {
            if(L[i] + len > R[i]) {
                flag = false;
                break;
            }
        }

        if(flag) return true;
        l = r + 1;
    }
    return false;
}

int main() {
    int T; scanf("%d", &T);
    while(T--) {
        scanf("%d", &m);

        char ch = 'A'; int tot = 0;
        for(int i = 1; i <= m; i++) {
            scanf("%s", str);
            int len = strlen(str);
            for(int j = 0; j < len; j++) {
                s[tot] = str[j];
                b[tot++] = i;
            }
            s[tot++] = ch++;
        }
        s[tot] = '\0';

        buildSa(s, tot + 1, 180);

        int l = 1, r = tot, mid, ans = 0;

        while(l <= r) {
            mid = l + r >> 1;
            if(check(mid, tot)) ans = mid, l = mid + 1;
            else r = mid - 1;
        }

        printf("%d\n", ans);
    }
    return 0;
}


/*
1
4
abbabba
dabddkababa
bacaba
baba
*/