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] - 1 - (i + 1) >= 0)
ret1 = query(L[i], i, 1) * query(i + 1, R[i] - 1, 1),
ret3 = query(L[i], i, 0) * query(i + 1, R[i] - 1, 0);
if (i - 1 - L[i] >= 0)
ret2 = query(L[i], i - 1, 1) * query(i, R[i] - 1, 1),
ret4 = query(L[i], i - 1, 0) * query(i, R[i] - 1, 0);
ret1 = max(ret1, ret3), ret2 = max(ret2, ret4);
ans2[het[i]] = max(max(ans2[het[i]], ret1), ret2);
}
ans1[0] = (LL(((LL)n * ((LL)n - 1LL)) / 2LL));
for (int i = 2; i < n; i++) {
LL ret1 = max(query(1, i, 1) * query(i + 1, n, 1),
query(1, i - 1, 1) * query(i, n, 1)),
ret2 = max(query(1, i, 0) * query(i + 1, n, 0),
query(1, i - 1, 0) * query(i, n, 0));
ans2[0] = max(max(ans2[0], ret1), ret2);
}
for (int i = n - 1; i >= 1; i--) {
ans1[i] = ans1[i + 1] + ans1[i];
if (ans2[i + 1] != 0) ans2[i] = max(ans2[i + 1], ans2[i]);
}
for (int i = 0; i < n; i++) if (ans2[i] == BASE) ans2[i] = 0;
for (int i = 0; i < n; i++) printf ("%lld %lld\n", ans1[i], ans2[i]);
}
int main() {
scanf("%d", &n); scanf ("%s", ss + 1);
for (int i = 1; i <= n; i++) scanf("%lld", &A[i]);
getSA(), prog(), getST();
solve1();
return 0;
}
