洛谷P2178 [NOI2015]品酒大会(后缀自动机 线段树)
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2022-03-26 15:01:00
题意 "题目链接" Sol 说一个后缀自动机+线段树的无脑做法 首先建出SAM,然后对parent树进行dp,维护最大次大值,最小次小值 显然一个串能更新答案的区间是$[len_{fa_{x}} + 1, len_x]$,方案数就相当于是从$siz_x$里面选两个,也就是$\frac{siz_x ( ......
题意
sol
说一个后缀自动机+线段树的无脑做法
首先建出sam,然后对parent树进行dp,维护最大次大值,最小次小值
显然一个串能更新答案的区间是\([len_{fa_{x}} + 1, len_x]\),方案数就相当于是从\(siz_x\)里面选两个,也就是\(\frac{siz_x (siz_x - 1)}{2}\)
直接拿线段树维护一下,标记永久化一下炒鸡好写~
#include<bits/stdc++.h> #define int long long #define ll long long using namespace std; const int maxn = 1e6 + 10; const ll inf = 2e18 + 10; template <typename a, typename b> inline bool chmin(a &a, b b){if(a > b) {a = b; return 1;} return 0;} template <typename a, typename b> inline bool chmax(a &a, b b){if(a < b) {a = b; return 1;} return 0;} inline int read() { char c = getchar(); int x = 0, f = 1; while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();} while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar(); return x * f; } int n, a[maxn]; char s[maxn]; int root = 1, tot = 1, las = 1, ch[maxn][26], fa[maxn], len[maxn], rev[maxn]; ll mx[maxn], mx2[maxn], ans[maxn], ans1[maxn], mn[maxn], mn2[maxn], tmp[maxn], siz[maxn]; vector<int> v[maxn]; void insert(int x, int id) { int now = ++tot, pre = las; las = now; siz[now] = 1; len[now] = len[pre] + 1; mx[now] = a[id]; mx2[now] = -inf; mn[now] = a[id]; mn2[now] = inf; rev[id] = now; for(; pre && !ch[pre][x]; pre = fa[pre]) ch[pre][x] = now; if(!pre) {fa[now] = root; return ;} int q = ch[pre][x]; if(len[pre] + 1 == len[q]) fa[now] = q; else { int nq = ++tot; fa[nq] = fa[q]; len[nq] = len[pre] + 1; memcpy(ch[nq], ch[q], sizeof(ch[q])); fa[now] = fa[q] = nq; for(; pre && ch[pre][x] == q; pre = fa[pre]) ch[pre][x] = nq; } } void builddag() { for(int i = 1; i <= tot; i++) assert(fa[i] != i), v[fa[i]].push_back(i); } int rt, node, ls[maxn], rs[maxn], ad[maxn], si[maxn]; ll sum[maxn], tag[maxn]; void build(int &k, int l, int r) { if(!k) k = ++node, tag[k] = -inf, si[k] = r - l + 1; if(l == r) return ; int mid = l + r >> 1; build(ls[k], l, mid); build(rs[k], mid + 1, r); } void intmax(int k, int l, int r, int ql, int qr, ll v) { if(ql <= l && r <= qr) {chmax(tag[k], v); return ; } int mid = l + r >> 1; if(ql <= mid) intmax(ls[k], l, mid, ql, qr, v); if(qr > mid) intmax(rs[k], mid + 1, r, ql, qr, v); } void intadd(int k, int l, int r, int ql, int qr, ll v) { if(ql <= l && r <= qr) {sum[k] += v; return ;} int mid = l + r >> 1; if(ql <= mid) intadd(ls[k], l, mid, ql, qr, v); if(qr > mid) intadd(rs[k], mid + 1, r, ql, qr, v); } ll querynum(int k, int l, int r, int pos) { if(!k) return 0; ll now = sum[k]; if(l == r || !k) return now; int mid = l + r >> 1; if(pos <= mid) now += querynum(ls[k], l, mid, pos); else now += querynum(rs[k], mid + 1, r, pos); return now; } ll querymax(int k, int l, int r, int pos) { if(!k) return -inf; ll now = tag[k]; if(l == r || !k) return now; int mid = l + r >> 1; if(pos <= mid) chmax(now, querymax(ls[k], l, mid, pos)); else chmax(now, querymax(rs[k], mid + 1, r, pos)); return now; } void dfs(int x) { for(auto &to : v[x]) { dfs(to); siz[x] += siz[to]; if(mx2[to] > mx[x]) chmax(mx2[x], mx[x]), mx[x] = mx2[to]; else chmax(mx2[x], mx2[to]); if(mx[to] > mx[x]) chmax(mx2[x], mx[x]), mx[x] = mx[to]; else chmax(mx2[x], mx[to]); if(mn2[to] < mn[x]) chmin(mn2[x], mn[x]), mn[x] = mn2[to]; else chmin(mn2[x], mn2[to]); if(mn[to] < mn[x]) chmin(mn2[x], mn[x]), mn[x] = mn[to]; else chmin(mn2[x], mn[to]); } if(siz[x] > 1 && x != root) { intmax(rt, 1, n, len[fa[x]] + 1, len[x], mx[x] * mx2[x]); intmax(rt, 1, n, len[fa[x]] + 1, len[x], mn[x] * mn2[x]); intadd(rt, 1, n, len[fa[x]] + 1, len[x], 1ll * siz[x] * (siz[x] - 1) / 2); } } signed main() { n = read(); build(rt, 1, n); scanf("%s", s + 1); reverse(s + 1, s + n + 1); for(int i = 1; i <= n; i++) tmp[i] = a[i] = read(), assert(a[i] != 0); reverse(a + 1, a + n + 1); for(int i = 1; i <= n; i++) insert(s[i] - 'a', i); for(int i = 1; i <= tot; i++) { ans[i] = -inf; if(!mx[i]) mx[i] = -inf; if(!mx2[i]) mx2[i] = -inf; if(!mn[i]) mn[i] = inf; if(!mn2[i]) mn2[i] = inf; } builddag(); dfs(1); for(int i = 1; i < n; i++) { ans1[i] = querynum(root, 1, n, i); ans[i] = querymax(root, 1, n, i); } sort(tmp + 1, tmp + n + 1, greater<int>()); cout << 1ll * n * (n - 1) / 2 << " " << max(tmp[1] * tmp[2], tmp[n] * tmp[n - 1]) << '\n'; for(int i = 1; i < n; i++) cout << ans1[i] << " " << (ans[i] <= -inf ? 0 : ans[i]) << '\n'; return 0; } /* 2 aa -100000000 100000000 12 abaabaabaaba 1 -2 3 -4 5 -6 7 -8 9 -10 11 -12 */
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