Codeforces 868 D. Huge Strings （二分+随机+SAM）

Description

You are given n strings s1, s2, …, sn consisting of characters 0 and 1. m operations are performed, on each of them you concatenate two existing strings into a new one. On the i-th operation the concatenation saisbi is saved into a new string sn + i (the operations are numbered starting from 1). After each operation you need to find the maximum positive integer k such that all possible strings consisting of 0 and 1 of length k (there are 2k such strings) are substrings of the new string. If there is no such k, print 0.

Input

The first line contains single integer n (1 ≤ n ≤ 100) — the number of strings. The next n lines contain strings s1, s2, …, sn (1 ≤ |si| ≤ 100), one per line. The total length of strings is not greater than 100.

The next line contains single integer m (1 ≤ m ≤ 100) — the number of operations. m lines follow, each of them contains two integers ai abd bi (1 ≤ ai, bi ≤ n + i - 1) — the number of strings that are concatenated to form sn + i.

Output

Print m lines, each should contain one integer — the answer to the question after the corresponding operation.

Example input

5
01
10
101
11111
0
3
1 2
6 5
4 4

Example output

1
2
0

题意

一个字符串最多可以包含几位二进制数的所有组合。

思路

以下做法属于旁门左道，二分 + 随机 + SAM 成功水过数据。

首先利用这个字符串建立后缀自动机，然后二分答案 k ，对于每一个 k 我们最多测试 1000 种组合，其中每一种组合都是随机生成的，然后在自动机中查找是否包含该子串。

因为初始的字符串可能是由某两个子串连接得到的，如果不做限制的话该串最长可以到达 2100×100 位。

我们假设当前字符串为 s ，它是由 a,b 连接得到的，假如 len(s)>1000 ， s=s.substr(0,500)+s.substr(len(s)−500,500) ，因为概率。。。

最终的结果为 max(test(a),test(b),test(s)) 。

AC 代码

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5+10;
const int maxm = 210;
typedef long long LL;
LL ans, K;

struct SAM
{
int ch[maxn][2];
int pre[maxn], step[maxn];
int last, id;
int nu[maxn];
void init()
{
last = id = 0;
memset(ch[0], -1, sizeof(ch[0]));
pre[0] = -1;
step[0] = 0;
}
void insert(int c)
{
int p = last, np = ++id;
step[np] = step[p] + 1;
memset(ch[np], -1, sizeof(ch[np]));

while (p != -1 && ch[p][c] == -1)
ch[p][c] = np, p = pre[p];
if (p == -1)
pre[np] = 0;
else
{
int q = ch[p][c];
if (step[q] != step[p] + 1)
{
int nq = ++id;
memcpy(ch[nq], ch[q], sizeof(ch[q]));
step[nq] = step[p] + 1;
pre[nq] = pre[q];
pre[np] = pre[q] = nq;
while (p != -1 && ch[p][c] == q)
ch[p][c] = nq, p = pre[p];
}
else
pre[np] = q;
}
last = np;
}
} sam;

unordered_map<string,bool> sk;
bitset<64> ss;

bool judge(const int &mid)
{
sk.clear();
LL maxx = min(1LL<<(mid+5),1000LL);
for(int i=0; i<maxx; i++)
{
ss = rand();
string now = ss.to_string();
if(sk[now])continue;
sk[now] = true;
int id = 0;
for(int j=0; j<mid; j++)
{
if(sam.ch[id][ss[j]]==-1)
return false;
id = sam.ch[id][ss[j]];
}
}
return true;
}

string s[maxm];
int n,m;
int dp[maxm];

int solve(int id)
{
sam.init();
int len = s[id].length();
for(int i=0; i<len; i++)
sam.insert(s[id][i]-'0');
int low = 0,high = 20;
while(true)
{
int mid = (low+high)>>1;
if(mid==low)
{
if(judge(high))
low = high;
break;
}
if(judge(mid))
low = mid;
else
high = mid-1;
}
return low;
}

int main()
{
srand((unsigned)time(NULL));
ios::sync_with_stdio(false);
cin.tie(0);
cin>>n;
for(int i=1; i<=n; i++)
cin>>s[i];
cin>>m;
for(int i=n+1; i<=n+m; i++)
{
int a,b;
cin>>a>>b;
s[i] = s[a]+s[b];
if(s[i].length()>1000)
s[i] = s[i].substr(0,500) + s[i].substr(s[i].length()-500,500);
dp[i] = max({dp[a],dp[b],solve(i)});
cout<<dp[i]<<endl;
}
return 0;
}