[CF369E]Valera and Queries_离线_树状数组

Valera and Queries

题目链接：codeforces.com/problemset/problem/369/E

数据范围：略。

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题解：

这种题，就单独考虑一次询问即可。

我们发现，包括了至少一个给定点的个数，等于总个数减掉一个给定点都不包括的线段数。

一个都不包括，就表示这个线段的在两个给定点中间，这个可以把线段抽象成二维平面上的点，然后离线+树状数组查询。

代码：

#include <bits/stdc++.h>

#define N 1000010 

using namespace std;

char *p1, *p2, buf[100000];

#define nc() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1 ++ )

int rd() {
	int x = 0;
	char c = nc();
	while (c < 48) {
		c = nc();
	}
	while (c > 47) {
		x = (((x << 2) + x) << 1) + (c ^ 48), c = nc();
	}
	return x;
}

int cnt = 0;

struct Node {
	int x, y1, y2, id, c, opt;
}a[N * 10];

inline bool cmp(const Node &a, const Node &b) {
	return a.x == b.x ? a.opt > b.opt : a.x < b.x;
}

int q[N];

int tree[N];

inline int lowbit(int x) {
	return x & (-x);
}

void update(int x) {
	for (int i = x; i < N; i += lowbit(i)) {
		tree[i] ++ ;
	}
}

int query(int x) {
	int ans = 0;
	for (int i = x; i; i -= lowbit(i)) {
		ans += tree[i];
	}
	return ans;
}

int ans[N];

int main() {
	int n = rd(), m = rd();
	// opt : 1 -> query, 2 -> update
	for (int i = 1; i <= n; i ++ ) {
		int x = rd(), y = rd();
		a[ ++ cnt] = (Node) {x, y, 0, 0, 1, 2};
	}
	// cout << cnt << endl ;
	for (int i = 1; i <= m; i ++ ) {
		int num = rd();
		for (int j = 1; j <= num; j ++ ) {
			q[j] = rd();
		}
		q[0] = 0;
		q[ ++ num] = N - 1;
		// cout << num << endl ;
		for (int j = 1; j <= num; j ++ ) {
			if (q[j] - q[j - 1] >= 2) {
				// printf("Fuck %d
", j);
				int x = q[j - 1] + 1, y = q[j] - 1;
				a[ ++ cnt] = (Node) {x - 1, x, y, i, -1, 1};
				a[ ++ cnt] = (Node) {y, x, y, i, 1, 1};
			}
		}
	}
	// cout << cnt << endl ;
	sort(a + 1, a + cnt + 1, cmp);
	// for (int i = 1; i <= cnt; i ++ ) {
	// 	printf("%d %d %d %d %d %d
", a[i].x, a[i].y1, a[i].y2, a[i].id, a[i].c, a[i].opt);
	// }
	for (int i = 1; i <= cnt; i ++ ) {
		// printf("id :: %d
", i);
		if (a[i].opt == 2) {
			update(a[i].y1);
		}
		else {
			// cout << query(a[i].y2) << ' ' << query(a[i].y1) << ' ' << a[i].c << endl ;
			// cout << (query(a[i].y2) - query(a[i].y1)) * a[i].c << endl ;
			ans[a[i].id] += (query(a[i].y2) - query(a[i].y1)) * a[i].c;
		}
	}

	for (int i = 1; i <= m; i ++ ) {
		printf("%d
", n - ans[i]);
	}
	return 0;
}