【HNOI2018】游戏

题面

题解

这道题目到底有没有靠谱一点的解法啊。。。

有很多种(color{green}{mathrm{AC}})的方法，设(L[i],R[i])表示点(i)最左边和最右边能够到达的位置

于是就有正着推(20)分，反着推(color{green}{mathrm{AC}})

还可以拓扑排序，正着加点(color{#001277}{mathrm{TLE}})，反着加点(color{green}{mathrm{AC}})

所以也没有什么好讲的了。

代码

#include<cstdio>
#include<cstring>
#include<cctype>
#include<algorithm>
#include<queue>
#define RG register
#define file(x) freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout)
#define clear(x, y) memset(x, y, sizeof(x))

inline int read()
{
	int data = 0, w = 1; char ch = getchar();
	while(ch != '-' && (!isdigit(ch))) ch = getchar();
	if(ch == '-') w = -1, ch = getchar();
	while(isdigit(ch)) data = data * 10 + (ch ^ 48), ch = getchar();
	return data * w;
}

const int maxn(1e6 + 10);
struct edge { int next, to; } e[maxn];
int head[maxn], e_num, deg[maxn];

inline void add_edge(int from, int to)
{
	e[++e_num] = (edge) {head[from], to};
	head[from] = e_num; ++deg[to];
}

int n, m, Q, p[maxn], cnt, key[maxn], L[maxn], R[maxn];
void TopSort()
{
	std::queue<int> Q;
	for(RG int i = n; i; i--) if(!deg[i]) Q.push(i);
	while(!Q.empty())
	{
		int x = Q.front(); Q.pop(); p[++cnt] = x;
		for(RG int i = head[x]; i; i = e[i].next)
			if(!--deg[e[i].to]) Q.push(e[i].to);
	}
}

void calc(int x)
{
	int l = x, r = x;
	while(1)
	{
		int pl = l, pr = r;
		while(l > 1 && (!key[l - 1] || (l <= key[l - 1] && key[l - 1] <= r)))
			l = L[l - 1];
		while(r < n && (!key[r] || (l <= key[r] && key[r] <= r))) r = R[r + 1];
		if(pl == l && pr == r) break;
	}
	L[x] = l, R[x] = r;
}

int main()
{
	n = read(), m = read(), Q = read();
	for(RG int i = 1, x, y; i <= m; i++)
	{
		x = read(), y = read(), key[x] = y;
		if(y <= x) add_edge(x + 1, x);
		else add_edge(x, x + 1);
	}
	TopSort();
	for(RG int i = 1; i <= n; i++) L[i] = R[i] = i;
	for(RG int i = 1; i <= n; i++) calc(p[i]);
	while(Q--)
	{
		int S = read(), T = read();
		puts(L[S] <= T && T <= R[S] ? "YES" : "NO");
	}
	return 0;
}