SPOJ - PHRASES

题意：

　　给n个字符串，求出最长的子串。使得子串在每个字符串中不重叠地至少出现2次。输出子串长度。

题解：

　　用后缀数组求出height数组，之后二分答案。check时对height数组进行分组，并维护每个字符串的最前和最后位置。

#include <bits/stdc++.h>

using namespace std;

const int N = 1e5+;

int t, n, k, len;

int vis[N], val[][];

char s[N];

int r[N];

int sa[N], t1[N], t2[N], c[N], height[N], rank[N];

void SA(int *r, int n, int m) {

    int *x = t1, *y = t2;

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

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

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

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

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

        int p = ;

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

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

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

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

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

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

        swap(x, y);

        p = ; x[sa[]] = ;

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

            x[sa[i]] = y[sa[i-]] == y[sa[i]] && y[sa[i-]+k] == y[sa[i]+k] ? p- : p++;

        if(p >= n) break;

        m = p;

    }

    int k = , j;

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

    for(int i = ; i < n; height[rank[i++]] = k)

        for(k ? k--:, j = sa[rank[i]-]; r[i+k] == r[j+k]; k++);

}

int add(int x, int v) {

    int p = vis[sa[x]], c = ;

    if(val[p][] == - && val[p][] == -) {

        val[p][] = val[p][] = sa[x];

        c++;

    }

    else {

        if(sa[x] < val[p][]) {

            if(val[p][] - val[p][] < v && val[p][] - sa[x] >= v) c++;

            val[p][] = sa[x];

        }

        else if(sa[x] > val[p][]) {

            if(val[p][] - val[p][] < v && sa[x] - val[p][] >= v) c++;

            val[p][] = sa[x];

        }

    }

    return c;

}

bool check(int x) {

    int cnt = ;

    int l = n;

    while(l < len) {

        if(height[l] < x) {

            memset(val, -, sizeof(val));

            cnt = ;

            cnt += add(l, x);

        }

        else cnt += add(l, x);

        l++;

        if(cnt == *n) return true;

    }

    return false;

}

int main() {

    scanf("%d", &t);

    while(t--) {

        len = ;

        scanf("%d", &n);

        memset(vis, , sizeof(vis));

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

            scanf("%s", s);

            k = strlen(s);

            for(int j = ; j < k; j++) r[len++] = s[j]-'a'+;

            vis[len] = ;

            r[len++] = i;

        }

        for(int i = ; i < len; i++) vis[i] += vis[i-];

        SA(r, len, );

        int l = , r = 1e5;

        while(l <= r) {

            int mid = l+r>>;

            if(check(mid)) l = mid+;

            else r = mid-;

        }

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

    }

}