【BZOJ 3669】【NOI 2014】魔法森林 LCT+枚举边

$LCT+枚举$ 复习一下$LCT$模板。

先以$Ai$为关键字$sort$，然后$Ai$从小到大枚举每条边，看能否构成环，构不成则加边，构成则判断，判断过了就切断$Bi$最大的边。

我的边是编号为$i+n$的点，忘了这点调了好久$QAQ$ $sosad$

#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 150003
#define read(x) x=getint()
using namespace std;
inline int getint() {int k = 0, fh = 1; char c = getchar();	for(; c < ‘0‘ || c > ‘9‘; c = getchar()) if (c == ‘-‘) fh = -1; for(; c >= ‘0‘ && c <= ‘9‘; c = getchar()) k = k * 10 + c - ‘0‘; return k * fh;}
struct nodeE {int a, b, x, y;} E[N];
struct node *null;
struct node {
	node *ch[2], *fa;
	int d, pos;
	short rev;
	bool pl() {return fa->ch[1] == this;}
	bool check() {return fa == null || (fa->ch[0] != this && fa->ch[1] != this);}
	void push() {if (rev) {rev = 0; swap(ch[0], ch[1]); ch[0]->rev ^= 1; ch[1]->rev ^= 1;}}
	void count() {
		pos = d;
		if (E[ch[0]->pos].b > E[pos].b) pos = ch[0]->pos;
		if (E[ch[1]->pos].b > E[pos].b) pos = ch[1]->pos;
	}
	void setc(node *r, bool c) {ch[c] = r; r->fa = this;}
} *rt[N];
node pool[N];
int n, m, tot = 0;
namespace LCT {
	int ans = 0x7fffffff;
	bool cmp(nodeE X, nodeE Y) {return X.a < Y.a;}
	node *newnode(int num = 0) {
		node *t = &pool[++tot];
		t->ch[0] = t->ch[1] = t->fa = null;
		t->d = t->pos = num; t->rev = 0;
		return t;
	}
	void Build() {
		null = &pool[0];
		null->ch[0] = null->ch[1] = null->fa = null;
		null->d = null->pos = null->rev = 0;
		read(n); read(m);
		for(int i = 1; i <= m; ++i)
			{read(E[i].x); read(E[i].y); read(E[i].a); read(E[i].b);}
		sort(E + 1, E + m + 1, cmp);
		for(int i = 1; i <= n; ++i)
			rt[i] = newnode();
		for(int i = 1; i <= m; ++i)
			rt[n + i] = newnode(i);
		E[0].b = 0;
	}
	void rotate(node *r) {
		node *f = r->fa;
		bool c = r->pl();
		if (f->check()) r->fa = f->fa;
		else f->fa->setc(r, f->pl());
		f->setc(r->ch[!c], c);
		r->setc(f, !c);
		f->count();
	}
	void update(node *r) {if (!r->check()) update(r->fa); r->push();}
	void splay(node *r) {
		update(r);
		for(; !r->check(); rotate(r))
			if (!r->fa->check()) rotate(r->pl() == r->fa->pl() ? r->fa : r);
		r->count();
	}
	node *access(node *r) {node *y = null;	for(; r != null; y = r, r = r->fa) {splay(r); r->ch[1] = y;} return y;}
	void changert(node *r) {access(r)->rev ^= 1; splay(r);}
	void link(node *r, node *t) {changert(r); r->fa = t;}
	void cut(node *r, node *t) {changert(r); access(t); splay(t); t->ch[0]->fa = null; t->ch[0] = null;}
	node *findrt(node *r) {access(r); splay(r); while(r->ch[0] != null) r = r->ch[0]; return r;}
	int ask(node *r, node *t) {changert(r); access(t); splay(t); return t->pos;}
	void work(int u, int v, int edge) {
		if (findrt(rt[u]) == findrt(rt[v])) {
			int k = ask(rt[u], rt[v]);
			if (E[k].b > E[edge - n].b) {
				cut(rt[u], rt[k + n]);
				cut(rt[v], rt[k + n]);
				link(rt[u], rt[edge]);
				link(rt[v], rt[edge]);
			}
		} else {
			link(rt[u], rt[edge]);
			link(rt[v], rt[edge]);
		}
		if (findrt(rt[1]) == findrt(rt[n]))
			ans = min(ans, E[edge - n].a + E[ask(rt[1], rt[n])].b);
	}
	void AC() {printf("%d\n", ans == 0x7fffffff ? -1 : ans);}
}

int main() {
	LCT::Build();
	for(int i = 1; i <= m ;++i)
		LCT::work(E[i].x, E[i].y, i + n);
	LCT::AC();
	return 0;
}

我的代码就是一堵墙，让$300$行的$LinkCutTree$在压行大法前颤抖吧～～～