洛谷P5111 zhtobu3232的线段树

题意：给定线段树，上面若干个节点坏了，求能表示出多少区间。

区间能被表示出当且仅当拆出来的log个节点都是好的。

解：每个区间在最浅的节点处计算答案。

对于每个节点维护从左边过来能有多少区间，从右边过来能有多少区间即可。

说起来水的一批为什么会被评成黑题啊。

 1 #include <bits/stdc++.h>
 2 
 3 typedef long long LL;
 4 const int N = 20000010;
 5 const LL MO = 998244353;
 6 
 7 LL lc[N], rc[N], ans;
 8 int tot, ls[N], rs[N];
 9 bool vis[N];
10 
11 void add(LL L, LL R, LL l, LL r, int &o) {
12     //printf("%lld %lld %lld %lld %d \n", L, R, l, r, o);
13     if(!o) o = ++tot;
14     if(L <= l && r <= R) {
15         vis[o] = 1;
16         return;
17     }
18     LL mid = (l + r) >> 1;
19     if((r - l + 1) & 1) mid--;
20     if(L <= mid) add(L, R, l, mid, ls[o]);
21     if(mid < R) add(L, R, mid + 1, r, rs[o]);
22 }
23 
24 void cal(LL l, LL r, int &o) {
25     if(!o) {
26         o = ++tot;
27         LL len = (r - l + 1) % MO;
28         ans = (ans + len * (len + 1) / 2) % MO;
29         //printf("[%lld %lld] ans += %lld = %lld \n", l, r, len * (len + 1) / 2, ans);
30         lc[o] = rc[o] = (len - 1 + MO) % MO;
31         return;
32     }
33     if(l == r) {
34         ans += (!vis[o]);
35         //if(!vis[o]) printf("[%lld %lld] ans++ = %lld \n", l, r, ans);
36         return;
37     }
38     LL mid = (l + r) >> 1;
39     if((r - l + 1) & 1) mid--;
40     cal(l, mid, ls[o]);
41     cal(mid + 1, r, rs[o]);
42     /// cal
43     ans = (ans + lc[rs[o]] * rc[ls[o]] % MO) % MO;
44     //printf("[%lld %lld] ans += %lld * %lld = %lld \n", l, r, lc[rs[o]], rc[ls[o]], ans);
45     if(!vis[ls[o]]) {
46         ans = (ans + lc[rs[o]]) % MO;
47         //printf("[%lld %lld] ans += %lld = %lld \n", l, r, lc[rs[o]], ans);
48     }
49     if(!vis[rs[o]]) {
50         ans = (ans + rc[ls[o]]) % MO;
51         //printf("[%lld %lld] ans += %lld = %lld \n", l, r, rc[ls[o]], ans);
52     }
53     if(!vis[o]) ans++;
54     lc[o] = lc[ls[o]];
55     if(!vis[ls[o]]) {
56         lc[o] = (lc[o] + lc[rs[o]] + 1) % MO;
57     }
58     rc[o] = rc[rs[o]];
59     if(!vis[rs[o]]) {
60         rc[o] = (rc[o] + rc[ls[o]] + 1) % MO;
61     }
62     //printf("l = %lld r = %lld lc = %lld rc = %lld \n", l, r, lc[o], rc[o]);
63     return;
64 }
65 
66 int main() {
67 
68     //printf("%d", sizeof(lc) * 3 / 1048576);
69 
70     LL n; int m, root = 0;
71     scanf("%lld%d", &n, &m);
72     for(int i = 1; i <= m; i++) {
73         LL l, r;
74         scanf("%lld%lld", &l, &r);
75         add(l, r, 1ll, n, root);
76     }
77     cal(1, n, root);
78     printf("%lld\n", ans);
79     return 0;
80 }