链接

Segment Game

3
0 0
0 3
0 1
5
0 1
0 0
1 1
0 1
0 0

Case #1:
0
0
0
Case #2:
0
1
0
2
Hint
For the second case in the sample:

At the first add operation,Lillian adds a segment [1,2] on the line.

At the second add operation,Lillian adds a segment [0,2] on the line.

At the delete operation,Lillian deletes a segment which added at the first add operation.

At the third add operation,Lillian adds a segment [1,4] on the line.

At the fourth add operation,Lillian adds a segment [0,4] on the line
 

每次插入一个线段，或删除一个已存在的线段。每次插入后输出当前插入的线段能完整覆盖存在的几条线段。

题解：对于新插入的线段，查询有多少个线段左端点大于等于该线段的左端点。 再查询有多少个线段的右端点大于该线段右端点， 两者之差就是答案。用两个树状数组搞定。时间复杂度nlogn

一共就4种情况。画绘图应该能发现。。

#pragma comment(linker, "/STACK:102400000,102400000")
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstring>
#include <queue>
#include <set>
#include <map>
#include <vector>
using namespace std;
template <class T>
inline bool rd(T &ret) {
	char c; int sgn;
	if (c = getchar(), c == EOF) return 0;
	while (c != ‘-‘ && (c<‘0‘ || c>‘9‘)) c = getchar();
	sgn = (c == ‘-‘) ? -1 : 1;
	ret = (c == ‘-‘) ?
 0 : (c - ‘0‘);
	while (c = getchar(), c >= ‘0‘&&c <= ‘9‘) ret = ret * 10 + (c - ‘0‘);
	ret *= sgn;
	return 1;
}
template <class T>
inline void pt(T x) {
	if (x < 0) {
		putchar(‘-‘);
		x = -x;
	}
	if (x > 9) pt(x / 10);
	putchar(x % 10 + ‘0‘);
}
typedef pair<int, int> pii;
typedef long long ll;
const int N = 450007;
struct Tree {
	int c[N], maxn;
	void init(int n) { maxn = n; for (int i = 0; i <= n; i++)c[i] = 0; }
	int lowbit(int x) { return x&-x; }
	int sum(int x) {
		int ans = 0;
		while (x)ans += c[x], x -= lowbit(x);
		return ans;
	}
	void update(int pos, int val) {
		while (pos <= maxn)c[pos] += val, pos += lowbit(pos);
	}
}A, B;
int n;
set<pii> s;
int op[N], l[N], r[N];
pii a[N];
vector<int>G;
int main() {
	int cas = 0;
	while (cin>>n) {
		G.clear();
		int top = 0;
		for (int i = 1; i <= n; i++) {
			rd(op[i]), rd(l[i]);
			if (op[i] == 0)
			{
				G.push_back(l[i]);
				r[i] = l[i] + (++top);
				G.push_back(r[i]);
			}
		}
		printf("Case #%d:\n", ++cas);
		sort(G.begin(), G.end()); G.erase(unique(G.begin(), G.end()), G.end());
		top = 0;
		for (int i = 1; i <= n; i++)
			if (op[i] == 0) {
				l[i] = lower_bound(G.begin(), G.end(), l[i]) - G.begin() + 1;
				r[i] = lower_bound(G.begin(), G.end(), r[i]) - G.begin() + 1;
				a[++top] = { l[i], r[i] };
			}
		A.init(G.size()); B.init(G.size());
		int all = 0;
		for (int i = 1; i <= n; i++)
		{
			if (op[i] == 0)
			{
				int ans = B.sum(r[i]);
				ans -= A.sum(l[i]-1);
				pt(ans); putchar(‘\n‘);
				A.update(l[i], 1);
				B.update(r[i], 1);
				all++;
			}
			else {
				A.update(a[l[i]].first, -1);
				B.update(a[l[i]].second, -1);
				all--;
			}
		}
	}
	return 0;
}
/*
99
7
1 2 2 1 3 !1 2
/*
1
????
0
1 8 3 7 2
*/