SPOJ PHRASES Relevant Phrases of Annihilation（后缀数组 + 二分）题解

题意：

给\(n\)个串，要你求出一个最长子串\(A\)，\(A\)在每个字串至少都出现\(2\)次且不覆盖，问\(A\)最长长度是多少

思路：

后缀数组处理完之后，二分这个长度，可以\(O(n)\)验证可行性，注意是“不覆盖”（英文不好看不懂），随便搞一下就好了。

代码：

#include<map>
#include<set>
#include<queue>
#include<cmath>
#include<stack>
#include<ctime>
#include<string>
#include<vector>
#include<cstdio>
#include<cstring>
#include<sstream>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 2e5 + 10;
const int INF = 0x3f3f3f3f;
const ull seed = 11;
const int MOD = 1e9 + 7;
using namespace std;

int str[maxn];
int t1[maxn], t2[maxn], c[maxn];
int sa[maxn];
int rk[maxn];
int height[maxn];
bool cmp(int *r, int a, int b, int l){
    return r[a] == r[b] && r[a + l] == r[b + l];
}
void da(int *str, int n, int m){
    n++;
    int i, j, p, *x = t1, *y = t2;
    for(i = 0; i < m; i++) c[i] = 0;
    for(i = 0; i < n; i++) c[x[i] = str[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(j = 1; j <= n; j <<= 1){
        p = 0;
        for(i = n - j; i < n; i++) y[p++] = i;
        for(i = 0; i < n; i++) if(sa[i] >= j) y[p++] = sa[i] - j;
        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(i = 1; i < n; i++)
            x[sa[i]] = cmp(y, sa[i - 1], sa[i], j)? p - 1 : p++;
        if(p >= n) break;
        m = p;
    }
    int k = 0;
    n--;
    for(i = 0; i <= n; i++) rk[sa[i]] = i;
    for(i = 0; i < n; i++){
        if(k) k--;
        j = sa[rk[i] - 1];
        while(str[i + k] == str[j + k]) k++;
        height[rk[i]] = k;
    }
}
char s[maxn];
int belong[maxn];
int num[15], pre[15];
int tot;
bool judge(int n){
    for(int i = 1; i <= n; i++){
        if(num[i] < 2) return false;
    }
    return true;
}
bool check(int len, int cnt, int n){
    memset(num, 0, sizeof(num));
    memset(pre, -1, sizeof(pre));
    for(int i = 2; i <= cnt; i++){
        if(height[i] >= len){
            if(pre[belong[sa[i - 1]]] == -1){
                num[belong[sa[i - 1]]]++;
                pre[belong[sa[i - 1]]] = sa[i - 1];
            }
            else if(abs(pre[belong[sa[i - 1]]] - sa[i - 1]) >= len){
                num[belong[sa[i - 1]]]++;
            }
            if(pre[belong[sa[i]]] == -1){
                num[belong[sa[i]]]++;
                pre[belong[sa[i]]] = sa[i];
            }
            else if(abs(pre[belong[sa[i]]] - sa[i]) >= len){
                num[belong[sa[i]]]++;
            }
        }
        else{
            if(judge(n)) return true;
            memset(num, 0, sizeof(num));
            memset(pre, -1, sizeof(pre));
        }
    }
    return judge(n);
}
int main(){
    int n;
    int T;
    scanf("%d", &T);
    while(T--){
        scanf("%d", &n);
        int cnt = 0;
        for(int i = 1; i <= n; i++){
            scanf("%s", s);
            int len = strlen(s);
            for(int j = 0; j < len; j++){
                belong[cnt] = i;
                str[cnt++] = s[j];
            }
            belong[cnt] = -i;
            str[cnt++] = i;
        }
        cnt--;
        str[cnt] = 0;
        da(str, cnt, 130);

        int l = 1, r = 10000;
        int Max = 0;
        while(l <= r){
            int m = (l + r) >> 1;
            if(check(m, cnt, n)){
                Max = m;
                l = m + 1;
            }
            else{
                r = m - 1;
            }
        }
        printf("%d\n", Max);
    }
    return 0;
}