题面
Sol
差分后就是求多个串的最长公共子串
套路啊
拼在一起用不同字符隔开，后缀数组，二分答案，分块height，开桶记录即可
我把差分值离散了
# include <bits/stdc++.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(2000000);
IL ll Read(){
RG char c = getchar(); RG ll x = 0, z = 1;
for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
return x * z;
}
int T, n, a[_], sa[_], rk[_], tmp[_], height[_], t[_], ans, mx, vis[_], cnt[1010], Index;
int o[_], num[1010][1010], len, l[_];
IL bool Cmp(RG int i, RG int j, RG int k){ return tmp[i] == tmp[j] && tmp[i + k] == tmp[j + k]; }
IL void Suffix_Sort(){
RG int m = mx;
for(RG int i = 1; i <= n; ++i) ++t[rk[i] = a[i]];
for(RG int i = 1; i <= m; ++i) t[i] += t[i - 1];
for(RG int i = n; i; --i) sa[t[rk[i]]--] = i;
for(RG int k = 1; k <= n; k <<= 1){
RG int l = 0;
for(RG int i = n - k + 1; i <= n; ++i) tmp[++l] = i;
for(RG int i = 1; i <= n; ++i) if(sa[i] > k) tmp[++l] = sa[i] - k;
for(RG int i = 0; i <= m; ++i) t[i] = 0;
for(RG int i = 1; i <= n; ++i) ++t[rk[tmp[i]]];
for(RG int i = 1; i <= m; ++i) t[i] += t[i - 1];
for(RG int i = n; i; --i) sa[t[rk[tmp[i]]]--] = tmp[i];
swap(rk, tmp); rk[sa[1]] = l = 1;
for(RG int i = 2; i <= n; ++i) rk[sa[i]] = Cmp(sa[i - 1], sa[i], k) ? l : ++l;
if(l >= n) break; m = l;
}
for(RG int i = 1, h = 0; i <= n; ++i){
if(h) --h;
while(a[i + h] == a[sa[rk[i] - 1] + h]) ++h;
height[rk[i]] = h;
}
}
IL bool Check(RG int x){
RG int tot = (vis[sa[1]] != 0); cnt[vis[sa[1]]] = ++Index;
for(RG int i = 2; i <= n; ++i){
if(height[i] >= x){
if(cnt[vis[sa[i]]] != Index) tot += (vis[sa[i]] != 0), cnt[vis[sa[i]]] = Index;
}
else cnt[vis[sa[i]]] = ++Index, tot = (vis[sa[i]] != 0);
if(tot == T) return 1;
}
return 0;
}
int main(RG int argc, RG char* argv[]){
T = Read();
for(RG int i = 1; i <= T; ++i){
l[i] = Read();
for(RG int j = 1; j <= l[i]; ++j) num[i][j] = Read();
}
for(RG int i = 1; i <= T; ++i)
for(RG int j = 1; j <= l[i]; ++j) o[++len] = num[i][j] = num[i][j + 1] - num[i][j];
sort(o + 1, o + len + 1); len = unique(o + 1, o + len + 1) - o - 1;
mx = len;
for(RG int i = 1; i <= T; ++i){
for(RG int j = 1; j <= l[i]; ++j){
num[i][j] = lower_bound(o + 1, o + len + 1, num[i][j]) - o;
a[++n] = num[i][j]; vis[n] = i;
}
a[++n] = ++mx;
}
Suffix_Sort();
Check(1);
RG int l = 0, r = n;
while(l <= r){
RG int mid = (l + r) >> 1;
if(Check(mid)) ans = mid + 1, l = mid + 1;
else r = mid - 1;
}
printf("%d
", ans);
return 0;
}