Card I space ....
This question should be the Chairman Tree entry question ... No modification, no discretization ... The people of the industry conscience AH
At the beginning of the blank tree I build out the result is a whole 2n more Node, and then the MLE ...
(double experience, see also)
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#include <cstdio>#include <cstring>#include <algorithm>#include <iostream> #define REP (i, n) for (int i = 0; i < n; ++i)#define CLR (x, C) memset (x, C, sizeof (x))#define M (L, R) (((L) + (R)) >> 1)using namespace std;const int MAXN = 500000 + 5;struct Node {Node *l, *r;int s;Node (): s (0) {L = R = this;}inline void Update () {s = L-S + R -S;}} pool[10700000], *pt = Pool, *root[MAXN];int v;node* Modify (node* t, int l, int r) {node* h = pt++;if (L = = r)H-s = t-S + 1;else {int m = m (l, R);if (v <= m) {H-L = Modify (T, L, L, m);h-r = t-r;} else {h-l = t, L;H-r = Modify (T-R, M + 1, R);}H-Update ();}return h;}int query (node* l, node* R, int l, int R) {if (L = = r) return l;int m = m (l, R);if (S > V, L-, S-l, L-R)return Query (L-L, R-L, L, m);else if (S > V, r, R, S-l)return Query (L-R, R-r, M + 1, R);return 0;}int main () {freopen ("test.in", "R", stdin);root[0] = pt++;int n, m;cin >> n >> m;Rep (i, N) {scanf ("%d", &v);root[i + 1] = Modify (root[i], 1, n);}While (m--) {int L, R;scanf ("%d%d", &l, &r);v = (r-l + 1) >> 1;printf ("%d\n", Query (root[l-1], root[R], 1, N));}return 0;}
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3524: [Poi2014]couriers time limit: Sec Memory Limit: MB
Submit: 782 Solved: 245
[Submit] [Status] [Discuss] Description
Give a sequence of length n. 1≤a[i]≤n.
The M-group asks, each time an interval [l,r] is asked, if there is a number that appears in [L,r] greater than (r-l+1)/2. If present, output this number, otherwise output 0.
Input
The first line is two numbers n,m.
Second row n number, A[i].
The next M-line, two numbers per line, l,r to ask [L,r] this interval.
Output
M lines, one answer per line.
Sample Input7 5
1 1 3 2 3 4 3
1 3
1 4
3 7
1 7
6 6
Sample Output1
0
3
0
4
HINT
"Data Range"
n,m≤500000
Source
by Dzy
Bzoj 3524: [Poi2014]couriers (Chairman tree)