题解 鼠

传送门

垃圾题解居然是 \(O(n\sqrt n\log n)\) 的

需要优化跳链
根号分治，使用光速幂状物
复杂度 \(O(n\sqrt n)\)

点击查看代码

// ubsan: undefined
// accoders
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 50010
#define ll long long
//#define int long long

char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
	int ans=0, f=1; char c=getchar();
	while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
	while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
	return ans*f;
}

int n, q;
int b[N], w[N];

namespace force{
	void solve() {
		for (int i=1,p,k,c; i<=q; ++i) {
			p=read(); k=read(); c=read();
			int step=0;
			while (p<=n) {
				if (step==k) c^=1, step=0;
				p=c?w[p]:b[p];
				++step;
			}
			printf("%d\n", p-n-1);
		}
	}
}

namespace task1{
	int f[2][N][226], g[2][N][226], h[2][N][226], sqr;
	void solve() {
		sqr=ceil(sqrt(n));
		for (int i=n+1; i<=n+2; ++i) {
			for (int j=0; j<=sqr; ++j) {
				f[0][i][j]=f[1][i][j]=i;
				g[0][i][j]=g[1][i][j]=i;
				h[0][i][j]=h[1][i][j]=i;
			}
		}
		for (int i=n; i; --i) {
			g[0][i][0]=g[1][i][0]=i;
			g[0][i][1]=b[i], g[1][i][1]=w[i];
			for (int j=2; j<=sqr; ++j) {
				g[0][i][j]=g[0][b[i]][j-1];
				g[1][i][j]=g[1][w[i]][j-1];
			}
			f[0][i][0]=f[1][i][0]=i;
			f[0][i][1]=g[0][i][sqr];
			f[1][i][1]=g[1][i][sqr];
			for (int j=2; j<=sqr; ++j) {
				f[0][i][j]=f[0][g[0][i][sqr]][j-1];
				f[1][i][j]=f[1][g[1][i][sqr]][j-1];
			}
			for (int j=1; j<=sqr; ++j) {
				h[0][i][j]=h[1][g[0][i][j]][j];
				h[1][i][j]=h[0][g[1][i][j]][j];
			}
		}
		for (int i=1,p,k,c; i<=q; ++i) {
			p=read(); k=read(); c=read();
			if (k<=sqr) printf("%d\n", h[c][p][k]-n-1);
			else {
				while (p<=n) {
					p=f[c][p][k/sqr];
					p=g[c][p][k%sqr];
					c^=1;
				}
				printf("%d\n", p-n-1);
			}
		}
	}
}

signed main()
{
	freopen("mouse.in", "r", stdin);
	freopen("mouse.out", "w", stdout);

	n=read(); q=read();
	for (int i=1; i<=n; ++i) b[i]=read(), w[i]=read();
	// force::solve();
	task1::solve();

	return 0;
}