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;
}