[BZOJ3669][Noi2014]魔法森林 && LCT

这道题有两个权值 我们把所有边按权值a排序 剩下的边都看成点放进一个LCT中 维护每一节点的最大权值点的位置 枚举所有的边 如果u, v连通 则删去最大的边 加入这条边

否则直接加入这条边 当发现1和n连通的时候更新答案

开数组的时候把val 和 MAX开成bool也是醉了orz

#include<cstdio>  
#include<algorithm>  
#include<cstring>  
#include<iostream>  
#include<queue>  
#define SF scanf  
#define PF printf  
using namespace std;  
typedef long long LL;
const int MAXN = 50000;
const int MAXM = 100000;
const int MAXNODE = 150000;
const int INF = 1234567890;
struct LCT {
	int ch[MAXNODE+10][2], fa[MAXNODE+10], MAX[MAXNODE+10];
	int val[MAXNODE+10], Q[MAXNODE+10];	
	bool rev[MAXNODE+10];
	bool isroot(int x) {
		return ch[fa[x]][0] != x && ch[fa[x]][1] != x;
	}
	void up(int x) {
		int ls = ch[x][0], rs = ch[x][1];
		MAX[x] = x;
		if(val[MAX[ls]] > val[MAX[x]]) MAX[x] = MAX[ls];
		if(val[MAX[rs]] > val[MAX[x]]) MAX[x] = MAX[rs];
	}
	void pushdown(int x) {
		int &ls = ch[x][0], &rs = ch[x][1];
		if(rev[x]) {
			rev[x] ^= 1; rev[ls] ^= 1; rev[rs] ^= 1;
			swap(ls, rs);
		}
	}
	void Rotate(int x) {
		int y = fa[x], z = fa[y];
		bool d = x == ch[y][0];
		if(!isroot(y)) ch[z][y == ch[z][1]] = x;
		fa[x] = z; fa[y] = x; fa[ch[x][d]] = y;
		ch[y][!d] = ch[x][d]; ch[x][d] = y;
		up(y);
	}
	void splay(int x) {
		int top = 0; Q[++top] = x;
		for(int i = x; !isroot(i); i = fa[i]) Q[++top] = fa[i];
		for(int i = top; i; i--) pushdown(Q[i]);
		while(!isroot(x)) {
			int y = fa[x];
			if(!isroot(y)) Rotate((ch[y][0] == x ^ ch[fa[y]][0] == y) ? x : y);
			Rotate(x);
		}
		up(x);
	}
	void access(int x) {
		int t = 0;
		while(x) {
			splay(x);
			ch[x][1] = t;
			t = x; x = fa[x];
		}
	}
	void makeroot(int x) {
		access(x); splay(x); rev[x] ^= 1;
	}
	void link(int x, int y) {
		makeroot(x); fa[x] = y; 
	}
	void cut(int x, int y) {
		makeroot(x); access(y); splay(y); ch[y][0] = fa[x] = 0;
	}
	int query(int x, int y) {
		makeroot(x); access(y); splay(y); return MAX[y];
	}
} sp;
int fa[MAXNODE+10], n, m, ans = INF;
int find(int x) {
	return fa[x] == x ? x : fa[x] = find(fa[x]);
}
struct Node {
	int u, v, a, b;
	bool operator < (const Node &t) const {
		return a < t.a;
	}
} E[MAXM+10];
int main() {
	SF("%d%d", &n, &m);
	for(int i = 1; i <= n; i++) fa[i] = i;
	for(int i = 1; i <= m; i++) SF("%d%d%d%d", &E[i].u, &E[i].v, &E[i].a, &E[i].b);
	sort(E+1, E+1+m);
	for(int i = 1; i <= m; i++) {
		sp.val[n+i] = E[i].b;
		sp.MAX[n+i] = n+i;
	}
	for(int i = 1; i <= m; i++) {
		int u = E[i].u, v = E[i].v, wt = E[i].b;
		int x = find(u), y = find(v);
		if(x != y) {
			fa[x] = y;
			sp.link(u, n+i); sp.link(v, n+i);
		}
		else {
			int t = sp.query(u, v);
			if(wt < sp.val[t]) {
				sp.cut(E[t-n].u, t); sp.cut(E[t-n].v, t);
				sp.link(u, n+i); sp.link(v, n+i);
			}
		}
		if(find(1) == find(n)) ans = min(ans, sp.val[sp.query(1, n)] + E[i].a);
	}
	PF("%d", ans == INF ? -1 : ans);
}