BZOJ4736 温暖会指引我们前行（LCT+最大生成树）

　　类似于瓶颈路，满足条件的路径一定在温度的最大生成树上，那么就是一个LCT维护MST的裸题了。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 100010
#define M 300010
#define inf 1000000001
#define lson tree[k].ch[0]
#define rson tree[k].ch[1]
#define lself tree[tree[k].fa].ch[0]
#define rself tree[tree[k].fa].ch[1]
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int n,m,cnt,w[N+M][2],id[M],temp[N+M];
struct data{int ch[2],fa,rev,len,sum,id;
}tree[N+M];
void up(int k)
{
    tree[k].sum=tree[lson].sum+tree[rson].sum+tree[k].len;
    tree[k].id=k;
    if (temp[tree[lson].id]<temp[tree[k].id]) tree[k].id=tree[lson].id;
    if (temp[tree[rson].id]<temp[tree[k].id]) tree[k].id=tree[rson].id;
}
void rev(int k){if (k) swap(lson,rson),tree[k].rev^=1;}
void down(int k){if (tree[k].rev) rev(lson),rev(rson),tree[k].rev=0;}
bool isroot(int k){return lself!=k&&rself!=k;}
int whichson(int k){return rself==k;}
void push(int k){if (!isroot(k)) push(tree[k].fa);down(k);}
void move(int k)
{
    int fa=tree[k].fa,gf=tree[fa].fa,p=whichson(k);
    if (!isroot(fa)) tree[gf].ch[whichson(fa)]=k;tree[k].fa=gf;
    tree[fa].ch[p]=tree[k].ch[!p],tree[tree[k].ch[!p]].fa=fa;
    tree[k].ch[!p]=fa,tree[fa].fa=k;
    up(fa),up(k);
}
void splay(int k)
{
    push(k);
    while (!isroot(k))
    {
        int fa=tree[k].fa;
        if (!isroot(fa))
            if (whichson(k)^whichson(fa)) move(k);
            else move(fa);
        move(k);
    }
}
void access(int k){for (int t=0;k;t=k,k=tree[k].fa) splay(k),tree[k].ch[1]=t,up(k);}
int findroot(int k){access(k),splay(k);for (;lson;k=lson) down(k);splay(k);return k;}
void makeroot(int k){access(k),splay(k),rev(k);}
void link(int x,int y){makeroot(x),tree[x].fa=y;}
void cut(int x,int y){makeroot(x),access(y),splay(y),tree[y].ch[0]=tree[x].fa=0,up(y);}
int queryid(int x,int y){makeroot(x),access(y),splay(y);return tree[y].id;}
int querylen(int x,int y){makeroot(x),access(y),splay(y);return tree[y].sum;}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4736.in","r",stdin);
    freopen("bzoj4736.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read(),m=read();
    for (int i=0;i<=n;i++) temp[i]=inf,tree[i].id=i;cnt=n;
    for (int i=1;i<=m;i++)
    {
        char c=getc();
        if (c=='f')
        {
            int u=read(),x=read()+1,y=read()+1,t=read(),l=read();
            if (findroot(x)==findroot(y))
            {
                int p=queryid(x,y);
                if (temp[p]>=t) continue;
                cut(p,w[p][0]),cut(p,w[p][1]);
                w[p][0]=w[p][1]=0;
            }
            id[u]=++cnt;temp[id[u]]=t,tree[id[u]].len=tree[id[u]].sum=l,tree[id[u]].id=id[u],w[id[u]][0]=x,w[id[u]][1]=y;
            link(id[u],x),link(id[u],y);
        }
        if (c=='m')
        {
            int x=read()+1,y=read()+1;
            if (findroot(x)!=findroot(y)) printf("-1\n");
            else printf("%d\n",querylen(x,y));
        }
        if (c=='c')
        {
            int u=read(),l=read();
            if (w[id[u]][0])
            {
                cut(id[u],w[id[u]][0]),cut(id[u],w[id[u]][1]);
                tree[id[u]].len=tree[id[u]].sum=l;
                link(id[u],w[id[u]][0]),link(id[u],w[id[u]][1]);
            }
        }
    }
    return 0;
}