传送门： http://www.lydsy.com/JudgeOnline/problem.php?id=2752
这道题其实想到线段树并不难，而且如果正确的推出了公式写起来也就是有点恶心而已＝ ＝实现不太容易出错。但是公式推起来确实比较的酸爽，加上我的公式又比较的迷，然后跑的飞慢＝ ＝
这道题对于一个区间的期望，是整个区间的路径和除上区间的路径数目。数目可以O(1), 然后就在线段树中维护这个路径和。考虑区间合并（线段树都是这样＝ ＝），那么对于两个区间l, r， 那么路径和首先等于l的路径和加上r路径和，然后对于l中的一段路径，那么它的更新就是这条路径左边点的个数(包括左端点）* r中点的个数 * v，而出了r中的点的个数以外，其余部分作为一个元素up进行维护。 同样右边同理一个down 维护。 同时在更新这两个值的时候还需要维护一个区间和sum＝ ＝
然后lazy 标记的下放：sum 直接搞，然后up 和down 都是 * len［区间长度］* (len - 1) * v;
ans += lazy * (1 * len * + 2 * (len - 1) + 3 * (len - 3) ... + (len - 1) * 2 + len * 1) = len * (len + 1) * (len + 2) / 6 * lazy
然后就搞搞＝ ＝
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; typedef long long qword; const qword maxn = 101000; struct node { qword up, down, ans, lazy, sum; node *l, *r; }e[maxn * 3]; qword ne = 0; node* root; node* build(qword l, qword r) { node* now = e + ne ++; if(l ^ r) { qword mid = (l + r) >> 1; now-> l = build(l, mid); now-> r = build(mid + 1, r); } return now; } void update(node* x, qword l, qword r, qword mid) { if(l ^ r) { x-> ans = x-> l-> ans + x-> r-> ans + x-> l-> up * (r - mid) + x-> r-> down * (mid - l + 1); x-> sum = x-> l-> sum + x-> r-> sum; x-> up = x-> l-> up + x-> r-> up + x-> r-> sum * (mid - l + 1); x-> down = x-> l-> down + x-> r-> down + x-> l-> sum * (r - mid); } } void pushdown(node* x, qword l, qword r) { if(x-> l) { qword mid = (l + r) >> 1, v = x-> lazy, ln = mid - l + 1, rn = r - mid; x-> l-> ans += ln * (ln + 1) / 2 * (ln + 2) / 3 * v; x-> r-> ans += rn * (rn + 1) / 2 * (rn + 2) / 3 * v; x-> l-> sum += ln * v; x-> r-> sum += rn * v; x-> l-> up += ln * (ln + 1) / 2 * v; x-> r-> up += rn * (rn + 1) / 2 * v; x-> l-> down += ln * (ln + 1) / 2 * v; x-> r-> down += rn * (rn + 1) / 2 * v; x-> l-> lazy += v, x-> r-> lazy += v; x-> lazy = 0; } } void addlazy(node* now, qword l, qword r, qword ls, qword rs, qword v) { if(now-> lazy != 0) pushdown(now, l, r); if(l == ls && r == rs) { qword nn = rs - ls + 1; now-> ans += nn * (nn + 1) / 2 * (nn + 2) / 3 * v; now-> sum += nn * v; now-> up += nn * (nn + 1) / 2 * v; now-> down += nn * (nn + 1) / 2 * v; now-> lazy += v; } else { qword mid = (l + r) >> 1; if(rs <= mid) addlazy(now-> l, l, mid, ls, rs, v); else if(ls > mid) addlazy(now-> r, mid + 1, r, ls, rs, v); else addlazy(now-> l, l, mid, ls, mid, v), addlazy(now-> r, mid + 1, r, mid + 1, rs, v); update(now, l, r, mid); } } node ask(node* now, qword l, qword r, qword ls, qword rs) { if(now-> lazy != 0) pushdown(now, l, r); node ret; if(l == ls && r == rs) ret = *now; else { qword mid = (l + r) >> 1; if(rs <= mid) ret = ask(now-> l, l, mid, ls, rs); else if(ls > mid) ret = ask(now-> r, mid + 1, r, ls, rs); else { node lr = ask(now-> l, l, mid, ls, mid); node rr = ask(now-> r, mid + 1, r, mid + 1, rs); ret.l = &lr, ret.r = &rr; update(&ret, ls, rs, mid); } } return ret; } qword qword_get() { qword x = 0; char c = (char)getchar(); qword f = 0; while(!isdigit(c) && c != '-') c = (char)getchar(); if(c == '-') c =(char)getchar(), f = 1; while(isdigit(c)) { x = x * 10 + (qword)(c - '0'); c = (char)getchar(); } if(f) x = -x; return x; } qword n, m; qword gcd(qword a, qword b) { return a % b == 0 ? b : gcd(b, a % b); } void test(node* now, qword l, qword r) { cout << l <<" "<< r <<" "<< now-> ans <<" "<< now-> sum <<" " << now-> up <<" "<< now-> down <<" "<< now-> lazy << endl; if(l ^ r) { qword mid = (l + r) >> 1; test(now-> l, l, mid), test(now-> r, mid + 1, r); } } void sov() { n = qword_get(), m = qword_get(); root = build(1, n - 1); char s[10]; while(m --) { scanf("%s", s + 1); if(s[1] =='C') { qword ls, rs, v; ls = qword_get(); rs = qword_get(); v = qword_get(); rs --; addlazy(root, 1, n - 1, ls, rs, v); } if(s[1] == 'Q') { qword ls, rs; ls = qword_get(), rs = qword_get(); rs --; node ret = ask(root, 1, n - 1, ls, rs); qword a = ret.ans; qword b = (rs - ls + 1) * (rs - ls + 2) / 2; qword g = gcd(a, b); printf("%lld/%lld\n", a / g, b / g); } } } int main() { //freopen("test.in", "r", stdin); //freopen("test.out", "w", stdout); sov(); return 0; }
