SPOJ

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+20;
int t, n, k, len;
int vis[N], val[15][2];
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 = 0; i < m; i++) c[i] = 0;
    for(int i = 0; i < n; i++) c[x[i] = r[i]]++;
    for(int i = 0; i < m; i++) c[i] += c[i-1];
    for(int i = n-1; i >= 0; i--) sa[--c[x[i]]] = i;
    for(int k = 1; k <= n; k <<= 1) {
        int p = 0;
        for(int i = n-k; i < n; i++) y[p++] = i; 
        for(int i = 0; i < n; i++) if(sa[i] >= k) y[p++] = sa[i]-k;
        for(int i = 0; i < m; i++) c[i] = 0;
        for(int i = 0; i < n; i++) c[x[y[i]]]++;
        for(int i = 0; i < m; i++) c[i] += c[i-1];
        for(int 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++) 
            x[sa[i]] = y[sa[i-1]] == y[sa[i]] && y[sa[i-1]+k] == y[sa[i]+k] ? p-1 : p++;
        if(p >= n) break;
        m = p;
    }
    int k = 0, j;
    for(int i = 0; i < n; i++) rank[sa[i]] = i;
    for(int i = 0; i < n; height[rank[i++]] = k)
        for(k ? k--:0, j = sa[rank[i]-1]; r[i+k] == r[j+k]; k++);
}
int add(int x, int v) {
    int p = vis[sa[x]], c = 0;
    if(val[p][0] == -1 && val[p][1] == -1) {
        val[p][0] = val[p][1] = sa[x];
        c++;
    }
    else {
        if(sa[x] < val[p][0]) {
            if(val[p][1] - val[p][0] < v && val[p][1] - sa[x] >= v) c++;
            val[p][0] = sa[x];
        }
        else if(sa[x] > val[p][1]) {
            if(val[p][1] - val[p][0] < v && sa[x] - val[p][0] >= v) c++;
            val[p][1] = sa[x];
        }
    }
    return c;
}
bool check(int x) {
    int cnt = 0;
    int l = n;
    while(l < len) {
        if(height[l] < x) {
            memset(val, -1, sizeof(val));
            cnt = 0;
            cnt += add(l, x);
        }
        else cnt += add(l, x);
        l++;
        if(cnt == 2*n) return true;
    }
    return false;
}
int main() {
    scanf("%d", &t);
    while(t--) {
        len = 0;
        scanf("%d", &n);
        memset(vis, 0, sizeof(vis));
        for(int i = 1; i <= n; i++) {
            scanf("%s", s);
            k = strlen(s);
            for(int j = 0; j < k; j++) r[len++] = s[j]-'a'+11;
            vis[len] = 1;
            r[len++] = i;
        }
        for(int i = 1; i < len; i++) vis[i] += vis[i-1];
        SA(r, len, 37);
        int l = 1, r = 1e5;
        while(l <= r) {
            int mid = l+r>>1;
            if(check(mid)) l = mid+1;
            else r = mid-1;
        }
        printf("%d
", r);
    }
}