BZOJ2001 HNOI2010城市建设（线段树分治+LCT）

一个很显然的思路是把边按时间段拆开线段树分治一下，用lct维护MST。理论上复杂度是O((M+Q)logNlogQ)，实际常数爆炸T成狗。正解写不动了。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<vector>
#include<stack>
using namespace std;
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;
}
#define N 20010
#define M 50010
#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]
int n,m,q,f[M],L[M<<2],R[M<<2],cnt;
long long tot=0,ans[M];
struct edge{int x,y,z;
}e[M],a[N+M<<1];
struct data{int i,x;};
vector<data> Q[M];
vector<int> tree[M<<2];
stack<int> undo[M<<2];
namespace lct
{
    struct data{int ch[2],fa,rev,x;}tree[N+M<<1];
    void up(int k)
    {
        tree[k].x=k;
        if (a[tree[lson].x].z>a[tree[k].x].z) tree[k].x=tree[lson].x;
        if (a[tree[rson].x].z>a[tree[k].x].z) tree[k].x=tree[rson].x;
    }
    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[fa].fa=k,tree[k].ch[!p]=fa;
        up(fa),up(k);
    }
    void splay(int k)
    {
        push(k);
        while (!isroot(k))
        {
            int fa=tree[k].fa;
            if (!isroot(fa))
                if (whichson(fa)^whichson(k)) 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);}
    void makeroot(int k){access(k),splay(k),rev(k);}
    int findroot(int k){access(k),splay(k);for (;lson;k=lson) down(k);splay(k);return 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 query(int x,int y){makeroot(x),access(y),splay(y);return tree[y].x;}
    void newpoint(int k){tot+=a[k].z;link(k,a[k].x),link(k,a[k].y);}
    void delpoint(int k){tot-=a[k].z;cut(k,a[k].x),cut(k,a[k].y);}
}
void build(int k,int l,int r)
{
    L[k]=l,R[k]=r;
    if (l==r) return;
    int mid=l+r>>1;
    build(k<<1,l,mid);
    build(k<<1|1,mid+1,r);
}
void add(int k,int l,int r,int e)
{
    if (L[k]==l&&R[k]==r) {tree[k].push_back(e);return;}
    int mid=L[k]+R[k]>>1;
    if (r<=mid) add(k<<1,l,r,e);
    else if (l>mid) add(k<<1|1,l,r,e);
    else add(k<<1,l,mid,e),add(k<<1|1,mid+1,r,e);
}
void solve(int k)
{
    int s=tree[k].size();
    for (int i=0;i<s;i++)
    if (lct::findroot(a[tree[k][i]].x)!=lct::findroot(a[tree[k][i]].y)) 
    lct::newpoint(tree[k][i]),undo[k].push(tree[k][i]);
    else 
    {
        int t=lct::query(a[tree[k][i]].x,a[tree[k][i]].y);
        if (a[t].z>a[tree[k][i]].z)
        {
            lct::delpoint(t),lct::newpoint(tree[k][i]);
            undo[k].push(-t),undo[k].push(tree[k][i]);
        }
    }
    if (L[k]==R[k]) ans[L[k]]=tot;
    else solve(k<<1),solve(k<<1|1);
    while (!undo[k].empty())
    {
        if (undo[k].top()<0) lct::newpoint(-undo[k].top());
        else lct::delpoint(undo[k].top());
        undo[k].pop();
    }
}
int main()
{
    freopen("bzoj2001.in","r",stdin);
    freopen("bzoj2001.out","w",stdout);
    n=read(),m=read(),q=read();
    for (int i=1;i<=m;i++) 
    e[i].x=read(),e[i].y=read(),e[i].z=read();
    for (int i=1;i<=q;i++)
    {
        int x=read(),y=read();
        Q[x].push_back((data){i,y});
    }
    build(1,1,q);
    cnt=n; 
    for (int i=1;i<=m;i++)
    {
        int s=Q[i].size();
        if (s==0) a[++cnt]=e[i],add(1,1,q,cnt);
        else
        {
            if (Q[i][0].i>1) a[++cnt]=e[i],add(1,1,Q[i][0].i-1,cnt);
            for (int j=0;j<s-1;j++)
            a[++cnt]=(edge){e[i].x,e[i].y,Q[i][j].x},
            add(1,Q[i][j].i,Q[i][j+1].i-1,cnt);
            a[++cnt]=(edge){e[i].x,e[i].y,Q[i][s-1].x},
            add(1,Q[i][s-1].i,q,cnt);
        }
    }
    solve(1);
    for (int i=1;i<=q;i++) printf("%lld
",ans[i]);
    fclose(stdin);fclose(stdout);
    return 0;
}