题目大意：给定一个n个点m条边的无向连通图，k次询问两点之间所有路径中最长边的最小值

LCT的裸题！首先维护一个动态的最小生成树，然后每次加入边时删除两点间路径上权值最大的边！最后询问时直接求x到y链上的最大权值即可！水爆了！！

。。。好吧开玩笑的 真正的题解见http://blog.csdn.net/popoqqq/article/details/39755703

我只是闲得无聊水一发LCT罢了0.0

TLE了好久。。。因为有边权为0的边我没更新。。。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define M 15010
#define INF 2147483647
using namespace std;
struct edge{
    int x,y,z;
}e[M<<1];
struct abcd{
    abcd *fa,*ls,*rs;
    int num,maxedge;
    bool rev_mark;
    abcd(int x);
    void Reverse();
    void Push_Up();
    void Push_Down();
}*null=new abcd(0),*tree[M],*edges[M<<1];
abcd :: abcd(int x)
{
    fa=ls=rs=null;
    num=maxedge=x;
    rev_mark=0;
}
void abcd :: Reverse()
{
    rev_mark^=1;
    swap(ls,rs);
}
void abcd :: Push_Up()
{
    int maxnum=-1;
    if(e[ls->maxedge].z>maxnum)
        maxnum=e[ls->maxedge].z,maxedge=ls->maxedge;
    if(e[rs->maxedge].z>maxnum)
        maxnum=e[rs->maxedge].z,maxedge=rs->maxedge;
    if(e[num].z>maxnum)
        maxedge=num;
     
}
void abcd :: Push_Down()
{
    if(fa->ls==this||fa->rs==this)
        fa->Push_Down();
    if(rev_mark)
    {
        ls->Reverse();
        rs->Reverse();
        rev_mark=0;
    }
}
void Zig(abcd *x)
{
    abcd *y=x->fa;
    y->ls=x->rs;
    x->rs->fa=y;
    x->rs=y;
    x->fa=y->fa;
    if(y==y->fa->ls)
        y->fa->ls=x;
    else if(y==y->fa->rs)
        y->fa->rs=x;
    y->fa=x;
    y->Push_Up();
}
void Zag(abcd *x)
{
    abcd *y=x->fa;
    y->rs=x->ls;
    x->ls->fa=y;
    x->ls=y;
    x->fa=y->fa;
    if(y==y->fa->ls)
        y->fa->ls=x;
    else if(y==y->fa->rs)
        y->fa->rs=x;
    y->fa=x;
    y->Push_Up();
}
void Splay(abcd *x)
{
    x->Push_Down();
    while(x->fa->ls==x||x->fa->rs==x)
    {
        abcd *y=x->fa,*z=y->fa;
        if(x==y->ls)
        {
            if(y==z->ls)
                Zig(y);
            Zig(x);
        }
        else
        {
            if(y==z->rs)
                Zag(y);
            Zag(x);
        }
    }
    x->Push_Up();
}
void Access(abcd *x)
{
    abcd *y=null;
    while(x!=null)
    {
        Splay(x);
        x->rs=y;
        x->Push_Up();
        y=x;
        x=x->fa;
    }
}
abcd* Find_Root(abcd *x)
{
    while(x->fa!=null)
        x=x->fa;
    return x;
}
void Move_To_Root(abcd *x)
{
    Access(x);
    Splay(x);
    x->Reverse();
}
void Link(abcd *x,abcd *y)
{
    Move_To_Root(x);
    x->fa=y;
}
void Cut(abcd *x,abcd *y)
{
    Move_To_Root(x);
    Access(y);
    Splay(y);
    x->fa=null;
    y->ls=null;
    y->Push_Up();
}
int Query(abcd *x,abcd *y)
{
    Move_To_Root(x);
    Access(y);
    Splay(y);
    return y->maxedge;
}
int n,m,k;
void Insert(abcd *x,abcd *y,int pos)
{
    if(e[pos].x==e[pos].y)
        return ;
    if( Find_Root(x)==Find_Root(y) )
    {
        int temp=Query(x,y);
        if(e[temp].z<=e[pos].z)
            return;
        Cut(edges[temp],tree[e[temp].x]);
        Cut(edges[temp],tree[e[temp].y]);
    }
    //if( Find_Root(x)==Find_Root(y) )
    //	printf("%d\n",1/0);
    Link(x,edges[pos]);
    Link(y,edges[pos]);
}
int main()
{
    int i,x,y;
    cin>>n>>m>>k;
    for(i=1;i<=n;i++)
        tree[i]=new abcd(0);
    for(i=1;i<=m;i++)
        edges[i]=new abcd(i);
    for(i=1;i<=m;i++)
    {
        scanf("%d%d%d",&e[i].x,&e[i].y,&e[i].z);
        Insert(tree[e[i].x],tree[e[i].y],i);
    }
    for(i=1;i<=k;i++)
    {
        scanf("%d%d",&x,&y);
        printf("%d\n", e[Query(tree[x],tree[y])].z );
    }
}

BZOJ 3732 Network Link-Cut-Tree (我是认真的！！