SPOJ220---Relevant Phrases of Annihilation（後綴數組+二分，對後綴分組）

題目
題意：給定N個串，求每個串至少出現兩次的最長子串
思路：二分枚舉長度，根據長度len分組，若某組裏的個數>=k，則說明存在長度爲len的至少重複k次子串。

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cstring>
using namespace std;

const int MAXN = 110000;

char s[MAXN];
int id[MAXN];
int sa[MAXN], rank[MAXN], height[MAXN], c[MAXN], tmp[MAXN];
int n, m, T;

void makesa(int m) {
    memset(c, 0, m * sizeof(int));
    for(int i = 0; i < n; ++i) ++c[rank[i] = s[i]];
    for(int i = 1; i < m; ++i) c[i] += c[i - 1];
    for(int i = 0; i < n; ++i) sa[--c[rank[i]]] = i;
    for(int k = 1; k < n; k <<= 1) {
        for(int i = 0; i < n; ++i) {
            int j = sa[i] - k;
            if(j < 0) j += n;
            tmp[c[rank[j]]++] = j;
        }
        int j = c[0] = sa[tmp[0]] = 0;
        for(int i = 1; i < n; ++i) {
            if(rank[tmp[i]] != rank[tmp[i - 1]] || rank[tmp[i] + k] != rank[tmp[i - 1] + k])
                c[++j] = i;
            sa[tmp[i]] = j;
        }
        memcpy(rank, sa, n * sizeof(int));
        memcpy(sa, tmp, n * sizeof(int));
    }
}

void calheight() {
    for(int i = 0, k = 0; i < n; height[rank[i++]] = k) {
        k -= (k > 0);
        int j = sa[rank[i] - 1];
        while(s[i + k] == s[j + k]) ++k;
    }
}

int mx[MAXN], mn[MAXN];
int stk[MAXN];

void update_max(int &a, int b) {
    if(a == -1 || a < b) a = b;
}

void update_min(int &a, int b) {
    if(a == -1 || a > b) a = b;
}

bool check(int L) {
    int sum = 0, top = 0;
    memset(mx, -1, m * sizeof(int));
    memset(mn, -1, m * sizeof(int));
    memset(c, 0, m * sizeof(int));
    for(int i = 0; i < n; ++i) {
        if(height[i] >= L) {
            update_max(mx[id[sa[i]]], sa[i]);
            update_min(mn[id[sa[i]]], sa[i]);
            stk[++top] = id[sa[i]];
            if(mx[id[sa[i]]] - mn[id[sa[i]]] >= L) {
                if(!c[id[sa[i]]]) ++sum;
                c[id[sa[i]]] = true;
                if(sum >= m) return true;
            }
        } else {
            sum = 0;
            while(top) {
                int t = stk[top--];
                mx[t] = mn[t] = -1;
                c[t] = false;
            }
            update_max(mx[id[sa[i]]], sa[i]);
            update_min(mn[id[sa[i]]], sa[i]);
            stk[++top] = id[sa[i]];
        }
    }
    return false;
}

int solve() {
    int l = 1, r = 5001;
    while(l < r) {
        int mid = (l + r) >> 1;
        if(check(mid)) l = mid + 1;
        else r = mid;
    }
    return l - 1;
}

int main() {
    scanf("%d", &T);
    while(T--) {
        scanf("%d", &m);
        n = 0;
        for(int i = 0; i < m; ++i) {
            scanf("%s", s + n);
            while(s[n]) id[n++] = i;
            s[n++] = i + 1;
        }
        s[n - 1] = 0;
        makesa(128);
        calheight();
        printf("%d\n", solve());
    }
}