解析

LCT从板子到算法的入门题吧
有一些不知道的很实用的技巧

把边按a排序从小到大加入边
那么我们只需要维护当前1-n路径上的b的最小值即可
如果这条边两端点本来不连通，就直接link
否则找到路径上b最大的一条边，断掉，再加入当前边（前提是b最大的边的b大于当前边）

实现上，可以把每条边转化成一个虚点
十分方便

代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N=2e5+100;
const int mod=1e9+7;
const double eps=1e-9;
inline ll read() {ll x(0),f(1);char c=getchar();while(!isdigit(c)) {if(c=='-')f=-1;c=getchar();}while(isdigit(c)) {x=(x<<1)+(x<<3)+c-'0';c=getchar();}return x*f;
}int n,m;#define ls(o) tr[o][0]
#define rs(o) tr[o][1]
int tr[N][2],f[N],rev[N],val[N],mx[N];
inline bool isroot(int x) {return tr[f[x]][0]!=x&&tr[f[x]][1]!=x;
}
inline bool which(int x) {return tr[f[x]][1]==x;
}
inline void pushup(int x) {if(x){mx[x]=x;if(ls(x)&&val[mx[ls(x)]]>val[mx[x]]) mx[x]=mx[ls(x)];if(rs(x)&&val[mx[rs(x)]]>val[mx[x]]) mx[x]=mx[rs(x)];}return;
}
inline void tag(int x) {if(x) {rev[x]^=1;swap(tr[x][0],tr[x][1]);}
}
inline void pushdown(int x) {if(rev[x]){rev[x]=0;tag(tr[x][0]);tag(tr[x][1]);}return;
}
void debug(int x) {if(!x) return;pushdown(x);debug(tr[x][0]);printf("debug: x=%d ls=%d rs=%d\n",x,tr[x][0],tr[x][1]);debug(tr[x][1]);return;
}
inline void rotate(int x) {int fa=f[x],gfa=f[fa];int d=which(x),son=tr[x][d^1];f[x]=gfa;if(!isroot(fa)) tr[gfa][which(fa)]=x;f[fa]=x;tr[x][d^1]=fa;if(son){f[son]=fa;}tr[fa][d]=son;pushup(fa);pushup(x);return;
}
int zhan[N];
inline void splay(int x) {int y=x,top=0;zhan[++top]=y;while(!isroot(y)) zhan[++top]=y=f[y];while(top) pushdown(zhan[top--]);for(int fa; fa=f[x],!isroot(x); rotate(x)) {if(!isroot(fa)) which(fa)==which(x)?rotate(fa):rotate(x);}return;
}
inline void access(int x) {for(int y(0); x; y=x,x=f[x]) {splay(x);tr[x][1]=y;pushup(x);if(y) f[y]=x;}return;
}
inline void makeroot(int x) {access(x);splay(x);tag(x);return;
}
inline int findroot(int x) {access(x);splay(x);while(pushdown(x),tr[x][0]) x=tr[x][0];splay(x);return x;
}
inline void link(int x,int y) {makeroot(x);if(findroot(y)==x) return;f[x]=y;pushup(y);//printf("link: %d -> %d\n",x,y);
}
inline void cut(int x,int y) {makeroot(x);access(y);splay(y);if(tr[y][0]!=x||tr[x][1]) return;tr[y][0]=f[x]=0;pushup(y);return;
}
inline void split(int x,int y){//y is fathermakeroot(x);access(y);splay(y);return;
}
struct edge{int x,y,a,b;bool operator < (const edge oth)const{return a<oth.a;}
}e[N];
int main() {
#ifndef ONLINE_JUDGE//freopen("a.in","r",stdin);//freopen("a.out","w",stdout);
#endifn=read();m=read();for(int i=1;i<=m;i++) e[i]=(edge){(int)read(),(int)read(),(int)read(),(int)read()};sort(e+1,e+1+m);int ans(2e9);for(int i=1;i<=m;i++){int x=e[i].x,y=e[i].y,w=e[i].b;val[i+n]=w;//printf("\nid=%d (%d %d) w=%d (a=%d)\n",i+n,x,y,w,e[i].a);if(findroot(x)!=findroot(y)) link(x,n+i),link(y,n+i);else{split(x,y);int now=mx[y];if(val[now]>w){splay(now);f[tr[now][0]]=0;tr[now][0]=0;f[tr[now][1]]=0;tr[now][1]=0;pushup(now);link(n+i,x);link(n+i,y);}}if(findroot(1)==findroot(n)){split(1,n);ans=min(ans,val[mx[n]]+e[i].a);//debug(n);}}printf("%d\n",ans<2e9?ans:-1);return 0;
}