【BZOJ5110】【CodePlus2017】Yazid 的新生舞会

【题目链接】

• 点击打开链接
  

【思路要点】

• 补档博客，无题解。

【代码】

#include
using namespace std;
#define MAXN	2000005
template  void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
struct Segment_Tree {
	struct Node {
		int lc, rc;
		long long sum, len, add, cnt;
		bool clear;
	} a[MAXN];
	int n, root, size;
	void build(int root, int l, int r) {
		a[root].len = r - l + 1;
		if (l == r) return;
		int mid = (l + r) / 2;
		a[root].lc = ++size;
		build(size, l, mid);
		a[root].rc = ++size;
		build(size, mid + 1, r);
	}
	void init(int x) {
		n = x;
		root = size = 0;
		build(root, 0, 2 * n);
	}
	long long func(long long len) {
		return len * (len - 1) / 2;
	}
	void pushdown(int root) {
		if (a[root].clear) {
			int tmp = a[root].lc;
			a[tmp].add = 0;
			a[tmp].sum = 0;
			a[tmp].cnt = 0;
			a[tmp].clear = true;
			tmp = a[root].rc;
			a[tmp].add = 0;
			a[tmp].sum = 0;
			a[tmp].cnt = 0;
			a[tmp].clear = true;
			a[root].clear = false;
		}
		if (a[root].add) {
			int tmp = a[root].lc;
			a[tmp].add += a[root].add;
			a[tmp].sum += a[tmp].len * a[root].add;
			tmp = a[root].rc;
			a[tmp].add += a[root].add;
			a[tmp].sum += a[tmp].len * a[root].add;
			a[root].add = 0;
		}
		if (a[root].cnt) {
			int tmp = a[root].lc;
			a[tmp].cnt += a[root].cnt;
			a[tmp].sum += func(a[tmp].len) * a[root].cnt;
			tmp = a[root].rc;
			a[tmp].cnt += a[root].cnt;
			a[tmp].add += a[root].cnt * a[a[root].lc].len;
			a[tmp].sum += func(a[tmp].len) * a[root].cnt + a[root].cnt * a[a[root].lc].len * a[tmp].len;
			a[root].cnt = 0;
		}
	}
	void update(int root) {
		a[root].sum = a[a[root].lc].sum + a[a[root].rc].sum;
	}
	void ModifyCnt(int root, int l, int r, int ql, int qr, long long d, long long extra) {
		if (l == ql && r == qr) {
			a[root].cnt += d;
			a[root].add += d * extra;
			a[root].sum += func(a[root].len) * d + extra * d * a[root].len;
			return;
		}
		pushdown(root);
		int mid = (l + r) / 2, len = 0;
		if (mid >= ql) ModifyCnt(a[root].lc, l, mid, ql, min(qr, mid), d, extra), len = min(qr, mid) - ql + 1;
		if (mid + 1 <= qr) ModifyCnt(a[root].rc, mid + 1, r, max(mid + 1, ql), qr, d, extra + len);
		update(root);
	}
	void ModifyAdd(int root, int l, int r, int ql, int qr, long long d) {
		if (l == ql && r == qr) {
			a[root].add += d;
			a[root].sum += a[root].len * d;
			return;
		}
		pushdown(root);
		int mid = (l + r) / 2;
		if (mid >= ql) ModifyAdd(a[root].lc, l, mid, ql, min(qr, mid), d);
		if (mid + 1 <= qr) ModifyAdd(a[root].rc, mid + 1, r, max(mid + 1, ql), qr, d);
		update(root);
	}
	void modify(int l, int r, int d) {
		if (l > r) return;
		l += n, r += n;
		ModifyCnt(root, 0, 2 * n, l, r, d, 0);
		ModifyAdd(root, 0, 2 * n, r + 1, 2 * n, (r - l + 1) * d);
	}
	long long query(int root, int l, int r, int ql, int qr) {
		if (l == ql && r == qr) return a[root].sum;
		pushdown(root);
		long long ans = 0;
		int mid = (l + r) / 2;
		if (ql <= mid) ans += query(a[root].lc, l, mid, ql, min(mid, qr));
		if (qr >= mid + 1) ans += query(a[root].rc, mid + 1, r, max(mid + 1, ql), qr);
		return ans;
	}
	long long query(int l, int r) {
		if (l > r) return 0;
		l += n, r += n;
		return query(root, 0, 2 * n, l, r);
	}
	void clear() {
		a[root].add = 0;
		a[root].cnt = 0;
		a[root].sum = 0;
		a[root].clear = true;
	}
} SGT;
vector  a[MAXN];
int n, type;
int main() {
	read(n), read(type);
	for (int i = 0; i <= n - 1; i++)
		a[i].push_back(0);
	int value = 0;
	for (int i = 1; i <= n; i++) {
		int x; read(x);
		a[x].push_back(i);
		value = max(value, x);
	}
	for (int i = 0; i <= value; i++)
		a[i].push_back(n + 1);
	SGT.init(n);
	long long ans = 0;
	for (int i = 0; i <= value; i++) {
		SGT.modify(-a[i][1] + 1, 0, 1);
		for (unsigned j = 1; j <= a[i].size() - 2; j++) {
			int tmp = -a[i][j] + 2 * j;
			ans += SGT.query(tmp, tmp);
			SGT.modify(tmp, tmp, 1);
			ans += SGT.query(tmp - (a[i][j + 1] - a[i][j]) + 1, tmp - 1);
			SGT.modify(tmp - (a[i][j + 1] - a[i][j]) + 1, tmp - 1, 1);
		}
		SGT.clear();
	}
	cout << ans << endl;
	return 0;
}