有nn个房间排成一列，编号为1,2,...,n1,2,...,n相邻的房间之间都有一道门。其中一部分们上锁（因此需要有对应的钥匙才能开门），其余的门都能直接打开。

现在小G告诉了小H每把锁的钥匙在哪个房间里（每把锁有且只有一把钥匙与之对应），并作出pp次指示：

第ii次让小H从第SiS_i个房间出发到TiT_i个房间里。但是小G有时会故意在指令中放入死路，而小H也不想浪费多余的体力去尝试，于是想事先调查清楚每次的指令是否会存在一条通路。

1≤n,p≤106,0≤m<n,1≤x,y,Si,Ti<n1\le n,p\le 10^6,0\le m <n,1\le x,y,S_i,T_i < n，保证xx不重复

Luogu P4436

Solution

容易发现，从一个房间开始后得到的这个可达区间一定是连续的一段，且一开始的锁把序列变成了几段，最暴力的想法是对于每一段，暴力地一左一右的扩展段

考虑优化掉一些重复转移，如果这个钥匙在这个门的右边，则只有右边的这段能扩展左边这段；否则右边的永远不可能更新到左边。即如果钥匙在门的右边，则先扩展左边这段再扩展右边；否则先扩展右边

按这个拓扑序暴力更新答案，复杂度O(n+m)O(n+m)

Code

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
#include <bits/stdc++.h>

#define x first
#define y second
#define y1 Y1
#define y2 Y2
#define mp make_pair
#define pb push_back

using namespace std;

typedef long long LL;
typedef pair<int, int> pii;

template <typename T> inline int Chkmax (T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline int Chkmin (T &a, T b) { return a > b ? a = b, 1 : 0; }

inline void proc_status()
{
	ifstream t ("/proc/self/status");
	cerr << string (istreambuf_iterator <char> (t), istreambuf_iterator <char> ()) <<endl;
}

inline int read ()
{
	int sum = 0, fl = 1; char ch = getchar();
	for (; !isdigit(ch); ch = getchar()) if (ch == '-') fl = -1;
	for (; isdigit(ch); ch = getchar()) sum = (sum << 3) + (sum << 1) + ch - '0';
	return sum * fl;
}

const int Maxn = 1e6 + 100;

int N, M, P, Pos[Maxn];
int L[Maxn], R[Maxn], Belong[Maxn];
int e, Begin[Maxn], To[Maxn << 1], Next[Maxn << 1], deg[Maxn];

inline void add_edge (int x, int y) { To[++e] = y; Next[e] = Begin[x]; Begin[x] = e; ++deg[y]; }

inline void Solve ()
{
	int cnt = 1; L[1] = R[1] = 1;
	for (int i = 2; i <= N; ++i)
	{
		if (Pos[i - 1])
		{
			L[++cnt] = i;
			if (Pos[i - 1] <= i - 1) add_edge (cnt, cnt - 1);
			else add_edge (cnt - 1, cnt);
		}
		R[cnt] = i;
	}

	for (int i = 1; i <= cnt; ++i) for (int j = L[i]; j <= R[i]; ++j) Belong[j] = i;

	static queue <int> Q;
	for (int i = 1; i <= cnt; ++i) if (!deg[i]) Q.push(i);
	while (!Q.empty())
	{
		int x = Q.front(), fl = 1; Q.pop();
		while (fl)
		{
			fl = 0;
			int pos1 = Pos[L[x] - 1], pos2 = Pos[R[x]];
			while (L[x] <= pos1 && pos1 <= R[x]) fl = 1, L[x] = L[Belong[L[x] - 1]], pos1 = Pos[L[x] - 1], pos2 = Pos[R[x]];
			while (L[x] <= pos2 && pos2 <= R[x]) fl = 1, R[x] = R[Belong[R[x] + 1]], pos2 = Pos[R[x]];
			while (L[x] <= pos1 && pos1 <= R[x]) fl = 1, L[x] = L[Belong[L[x] - 1]], pos1 = Pos[L[x] - 1], pos2 = Pos[R[x]];
		}
		for (int i = Begin[x]; i; i = Next[i])
		{
			int y = To[i];
			--deg[y];
			if (!deg[y]) Q.push(y);
		}
	}

//	for (int i = 1; i <= cnt; ++i) cout<<L[i]<<" "<<R[i]<<endl;

	while (P--)
	{
		int x = read(), y = read();
		if (L[Belong[x]] <= y && y <= R[Belong[x]]) puts("YES");
		else puts("NO");
	}
}

inline void Input ()
{
	N = read(), M = read(), P = read();
	for (int i = 1; i <= M; ++i)
	{
		int x = read(), y = read();
		Pos[x] = y;
	}
}

int main()
{
#ifdef hk_cnyali
	freopen("A.in", "r", stdin);
	freopen("A.out", "w", stdout);
#endif
	Input();
	Solve();
	return 0;
}