洛谷 P2542 [AHOI2005]航线规划（Link-cut-tree）

题面

洛谷

bzoj

题解

离线处理+LCT

有点像星球大战

我们可以倒着做，断边变成连边

我们可以把边变成一个点

连边时，如果两个点本身不联通，就\(val\)赋为\(1\)，并连接这条边

如果，两个点本身就联通，那么就不连接这条边，把两点之间的\(val\)全部赋为\(0\)

\(ans\)就是求两点之间的\(sum(val)\)

Code

#include<bits/stdc++.h>

#define LL long long
#define RG register

using namespace std;

inline int gi() {
	RG int x = 0; RG char c = getchar(); bool f = 0;
	while (c != '-' && (c < '0' || c > '9')) c = getchar();
	if (c == '-') c = getchar(), f = 1;
	while (c >= '0' && c <= '9') x = x*10+c-'0', c = getchar();
	return f ? -x : x;
}

const int N = 52000;
struct node {
	int ch[2], f, sum, rev, v, ly;
}t[N<<2];
bool isroot(int x) {
	return t[t[x].f].ch[0] != x && t[t[x].f].ch[1] != x;
}
inline int get(int x) {
	return t[t[x].f].ch[1] == x;
}
inline void pushup(int x) {
	t[x].sum = t[t[x].ch[0]].sum+t[t[x].ch[1]].sum+t[x].v;
}
void rotate(int x) {
	int y = t[x].f, z = t[y].f, k = get(x);
	if (!isroot(y))
		t[z].ch[get(y)] = x;
	t[x].f = z;
	t[t[x].ch[k^1]].f = y; t[y].ch[k] = t[x].ch[k^1];
	t[y].f = x; t[x].ch[k^1] = y;
	pushup(y);
	return ;
}
int top, S[N<<2];
inline void putly(int x) {t[x].ly = 1; t[x].sum = t[x].v = 0;}
void pushdown(int x) {
	if (t[x].rev) {
		swap(t[x].ch[0], t[x].ch[1]);
		if (t[x].ch[0]) t[t[x].ch[0]].rev ^= 1;
		if (t[x].ch[1]) t[t[x].ch[1]].rev ^= 1;
		t[x].rev = 0;
	}
	if (t[x].ly) {
		if (t[x].ch[0]) putly(t[x].ch[0]);
		if (t[x].ch[1]) putly(t[x].ch[1]);
		t[x].ly = 0;
	}
	return ;
}
void splay(int x) {
	S[top=1] = x;
	for (int i = x; !isroot(i); i = t[i].f) S[++top] = t[i].f;
	for (int i = top; i; i--) pushdown(S[i]);
	while (!isroot(x)) {
		int y = t[x].f;
		if (!isroot(y))(get(x)^get(y)) ? rotate(x):rotate(y);
		rotate(x);
	}
	pushup(x);
	return ;
}
void access(int x) {for (int y=0;x;y=x,x=t[x].f) splay(x),t[x].ch[1]=y,pushup(x);}
void makeroot(int x){access(x),splay(x),t[x].rev ^= 1;}
void split(int x, int y){makeroot(x),access(y),splay(y);}
void link(int x, int y) {makeroot(x);t[x].f = y;}
int findroot(int x) {access(x); splay(x);while (t[x].ch[0]) x = t[x].ch[0];return x;}

struct Line {
	int u, v;
	bool flag;
}E[N<<2];
map<pair<int, int>, int> M;
struct question {
	int op, ans, u, v;
}q[N];

int main() {
	//freopen(".in", "r", stdin);
	//freopen(".out", "w", stdout);
	int n = gi(), m = gi();
	for (int i = 1; i <= m; i++) {
		int u = gi(), v = gi();
		if (u > v) swap(u, v);
		E[i] = (Line) {u, v, 0};
		t[i+n].v = 1;
		M[make_pair(u, v)] = i;
	}
	int cnt = 0;
	for (;;) {
		int c = gi(); if (c == -1) break;
		int u = gi(), v = gi(); if (u > v) swap(u, v);
		q[++cnt] = (question) {c, 0, u, v};
		if (!c) E[q[cnt].ans = M[make_pair(u, v)]].flag = 1;
	}
	for (int i = 1; i <= m; i++)
		if (!E[i].flag) {
			int u = E[i].u, v = E[i].v;
			if (findroot(u) != findroot(v))
				link(u, i+n), link(v, i+n);
			else {
				split(u, v);
				putly(v);
			}
		}
	for (int i = cnt; i; i--) {
		if (q[i].op) {
			int u = q[i].u, v = q[i].v;
			split(u, v);
			q[i].ans = t[v].sum;
		}
		else {
			int u = q[i].u, v = q[i].v;
			if (findroot(u) != findroot(v))
				link(u, q[i].ans+n), link(v, q[i].ans+n);
			else {
				split(u, v);
				putly(v);
			}
		}
	}
	for (int i = 1; i <= cnt; i++)
		if (q[i].op)
			printf("%d\n", q[i].ans);
	return 0;
}