BZOJ 4199 品酒大会
题意：
给定一个长度为nn的串。n≤3×105n≤3×105
对于∀i∈[0,n−1]∀i∈[0,n−1]，求出LCP(suf(k),suf(j))==iLCP(suf(k),suf(j))==i的数量并且求出其最大的权值积。
https://www.lydsy.com/JudgeOnline/problem.php?id=4199
Sol：
单调栈求出每个Height[i]Height[i]能延伸的范围，然后乘起来，注意Height[i]=0Height[i]=0时不能这么求
原因是会有多个00，每个00都会控制最小左边界和最大右边界，也就是说会算多次但是不对。
第二问直接STST表然后对于每个点ii求左右的最大最小然后匹配一下
数据比较弱，明显都没卡对于一个ii，这个ii是划分到左区间还是划分到右区间。
upd ： 并查集做法好像 快很多的样子，好像也很好写MergeMerge有点烦
#include<cctype>
#include<cstdio> #include<cstring> #include<algorithm> #include<algorithm> const int N = 1e5 + 5e4 + 7; const int lim = 3e5; typedef long long LL; inline LL max(LL a, LL b){return a > b ? a : b;} inline LL min(LL a, LL b){return a > b ? b : a;} int x[N*2], y[N*2], sa[N*2], rk[N*2], c[N*2]; int m = lim, n; int het[N*2], qlog[N*2];LL st[20][N*2][2]; //#define R register char ss[N*2]; inline void getSA() { for (int i = 0; i <= m; i++) c[i] = 0; for (int i = 1; i <= n; i++) c[x[i] = ss[i] - 'a' + 1]++; for (int i = 1; i <= m; i++) c[i] += c[i - 1]; for (int i = 1; i <= n; i++) sa[c[x[i]]--] = i; for (int siz = 1, p = 0; siz <= n; siz <<= 1, p = 0) { for (int i = n - siz + 1; i <= n; i++) y[++p] = i; for (int i = 1; i <= n; i++) if (sa[i] > siz) y[++p] = sa[i] - siz; for (int i = 0; i <= m; i++) c[i] = 0; for (int i = 1; i <= n; i++) c[x[y[i]]]++; for (int i = 1; i <= m; i++) c[i] += c[i - 1]; for (int i = n; i >= 1; i--) sa[c[x[y[i]]]--] = y[i], y[i] = 0; std :: swap(x, y), x[sa[1]] = 1, p = 1; for (int i = 2; i <= n; i++) x[sa[i]] = (y[sa[i]] == y[sa[i - 1]] && y[sa[i] + siz] == y[sa[i - 1] + siz]) ? p : ++p; if (p == n) break; m = p; } int k = 0; memset(het, 0, sizeof(het)); for (int i = 1; i <= n; i++) rk[sa[i]] = i; for (int i = 1; i <= n; i++) if (rk[i] != 1) { if (k) k--; int j = sa[rk[i] - 1]; while(j + k <= n && i + k <= n && ss[i + k] == ss[j + k]) k++; het[rk[i]] = k; } } LL inf = (LL)(1e9 + 7); LL A[N*2]; inline void getST() { qlog[1] = 0, memset(st, 0, sizeof(st)); for (int i = 2; i <= lim; i++) qlog[i] = qlog[i / 2] + 1; for (int i = 1; i <= n; i++) st[0][i][0] = st[0][i][1] = A[sa[i]]; // 0 最大 1最小 int logw = qlog[n]; for (int i = 1; i <= logw; i++) for (int j = 1; j + (1 << i) - 1 <= n; j++) { st[i][j][0] = max(st[i - 1][j][0], st[i - 1][j + (1 << (i - 1))][0]); st[i][j][1] = min(st[i - 1][j][1], st[i - 1][j + (1 << (i - 1))][1]); } } LL BASE = -(inf * inf); inline LL query(int x, int y, int cal) { LL ans1 = BASE, ans2 = BASE; if (x > y) return 1; else if (x == y) return A[sa[x]]; int logw = qlog[y - x + 1]; ans1 = max(st[logw][x][0], st[logw][y - (1 << logw) + 1][0]); ans2 = min(st[logw][x][1], st[logw][y - (1 << logw) + 1][1]); return cal ? ans1 : ans2; } int stack[N*2], top;int R[N*2], L[N*2]; void prog() { stack[++top] = 1; for (int i = 2; i <= n; i++) { while (top && het[stack[top]] >= het[i]) R[stack[top--]] = i; L[i] = stack[top], stack[++top] = i; } while (top) R[stack[top--]] = n + 1; } LL ans1[2*N]; LL ans2[2*N]; inline void solve1() { LL res = -(LL)(inf * inf); for (int i = 0; i <= n; i++) ans2[het[i]] = res; for (int i = 2; i <= n; i++) { if (!het[i]) continue; ans1[het[i]] += (LL)((LL)R[i] - i) * ((LL)i - L[i]); if (R[i] - L[i] <= 1) continue; LL ret1, ret2, ret3, ret4; ret1 = ret2 = ret3 = ret4 = BASE; if (R[i] -