P4069 [SDOI2016]游戏

题意

显然书剖套李超树。

考虑怎么算函数值：

设\((x,y)\)的\(lca\)为\(z\)，我们插一条斜率为\(k\)，截距为\(b\)的线段。

\((x,z)\)上的点\(u\)：
\(f(u)=k*(dis[x]-dis[u])+b=-k*dis[u]+(k*dis[x]+b)\)
所以对这条路径插入斜率为\(-k\)，截距为\(k*dis[x]+b\)的线段

\((y,z)\)上的点\(u\)：
\(f(u)=k*(dis[y]+dis[u]-2*dis[z])+b=k*dis[u]+k*(dis[y]-2*dis[z])+b\)
同理

code:

#include<bits/stdc++.h>
using namespace std;
#define int long long
#define ls(p) (p<<1)
#define rs(p) (p<<1|1)
#define minn(p)(seg[p].minn)
#define minid(p) (seg[p].minid)
const int maxn=100010;
const int inf=123456789123456789ll;
int n,m,cnt,tim;
int head[maxn],dep[maxn],dfn[maxn],pos[maxn],size[maxn],son[maxn],pre[maxn],top[maxn],dis[maxn];
struct edge{int to,nxt,dis;}e[maxn<<1];
struct Line
{
    int k,b;
    inline int calc(int x){return k*x+b;}
};
struct Seg{int minn;Line minid;}seg[maxn<<2];
inline int read()
{
    char c=getchar();int res=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9')res=res*10+c-'0',c=getchar();
    return res*f;
}
inline void add(int u,int v,int w)
{
    e[++cnt].nxt=head[u];
    head[u]=cnt;
    e[cnt].to=v;
    e[cnt].dis=w;
}
void dfs1(int x,int fa)
{
    pre[x]=fa;dep[x]=dep[fa]+1;size[x]=1;
    for(int i=head[x];i;i=e[i].nxt)
    {
        int y=e[i].to;
        if(y==fa)continue;
        dis[y]=dis[x]+e[i].dis;
        dfs1(y,x);size[x]+=size[y];
        if(size[son[x]]<size[y])son[x]=y;
    }
}
void dfs2(int x,int tp)
{
    top[x]=tp;dfn[x]=++tim;pos[tim]=x;
    if(son[x])dfs2(son[x],tp);
    for(int i=head[x];i;i=e[i].nxt)
    {
        int y=e[i].to;
        if(y==pre[x]||y==son[x])continue;
        dfs2(y,y);
    }
}
inline int lca(int x,int y)
{
    while(top[x]!=top[y])
    {
        if(dep[top[x]]<dep[top[y]])swap(x,y);
        x=pre[top[x]];
    }
    if(dep[x]>dep[y])swap(x,y);
    return x;
}
inline void up(int p){minn(p)=min(minn(p),min(minn(ls(p)),minn(rs(p))));}
void build(int p,int l,int r)
{
    minn(p)=inf;minid(p)=(Line){0,inf};
    if(l==r)return;
    int mid=(l+r)>>1;
    build(ls(p),l,mid);build(rs(p),mid+1,r);
}
inline void move(int p,int l,int r,Line id)
{
    int nowl=minid(p).calc(dis[pos[l]]),nowr=minid(p).calc(dis[pos[r]]);
    int tmpl=id.calc(dis[pos[l]]),tmpr=id.calc(dis[pos[r]]);
    if(nowl<=tmpl&&nowr<=tmpr)return;
    if(nowl>=tmpl&&nowr>=tmpr)
    {
        minid(p)=id;minn(p)=min(minn(p),min(tmpl,tmpr));
        return;
    }
    double point=1.0*(minid(p).b-id.b)/(1.0*(id.k-minid(p).k));
    int mid=(l+r)>>1;
    if(tmpl>nowl)
    {
        if(point<=(double)dis[pos[mid]])move(ls(p),l,mid,minid(p)),minid(p)=id;
        else move(rs(p),mid+1,r,id);
    }
    else
    {
        if(point<=(double)dis[pos[mid]])move(ls(p),l,mid,id);
        else move(rs(p),mid+1,r,minid(p)),minid(p)=id;
    } 
    minn(p)=min(minn(p),min(tmpl,tmpr));
    up(p);
}
void change(int p,int l,int r,int ql,int qr,Line id)
{
    if(l>=ql&&r<=qr){move(p,l,r,id);return;}
    int mid=(l+r)>>1;
    if(ql<=mid)change(ls(p),l,mid,ql,qr,id);
    if(qr>mid)change(rs(p),mid+1,r,ql,qr,id);
    up(p);
}
int query(int p,int l,int r,int ql,int qr)
{
    if(l>=ql&&r<=qr)return minn(p);
    int mid=(l+r)>>1,res=min(minid(p).calc(dis[pos[max(l,ql)]]),minid(p).calc(dis[pos[min(r,qr)]]));
    if(ql<=mid)res=min(res,query(ls(p),l,mid,ql,qr));
    if(qr>mid)res=min(res,query(rs(p),mid+1,r,ql,qr));
    return res; 
}
inline void trchange(int x,int y,int k,int b)
{
    while(top[x]!=top[y])
    {
        change(1,1,n,dfn[top[x]],dfn[x],(Line){k,b});
        x=pre[top[x]];
    }
    change(1,1,n,dfn[y],dfn[x],(Line){k,b});
}
inline void trsolve(int x,int y,int k,int b)
{
    int z=lca(x,y);
    trchange(x,z,-k,k*dis[x]+b);
    trchange(y,z,k,k*(dis[x]-(dis[z]<<1))+b);
}
inline int trquery(int x,int y)
{
    int res=inf;
    while(top[x]!=top[y])
    {
        if(dep[top[x]]<dep[top[y]])swap(x,y);
        res=min(res,query(1,1,n,dfn[top[x]],dfn[x]));
        x=pre[top[x]];
    }
    if(dep[x]>dep[y])swap(x,y);
    res=min(res,query(1,1,n,dfn[x],dfn[y]));
    return res;
}
signed main()
{
    //freopen("test.in","r",stdin);
    //freopen("test.out","w",stdout);
    n=read(),m=read();
    for(int i=1;i<n;i++)
    {
        int u=read(),v=read(),w=read();
        add(u,v,w),add(v,u,w);
    }
    dfs1(1,0),dfs2(1,1);
    build(1,1,n);
    for(int i=1;i<=m;i++)
    {
        int op=read(),x=read(),y=read(),k,b;
        if(op==1)k=read(),b=read(),trsolve(x,y,k,b);
        else printf("%lld\n",trquery(x,y));
    }
    return 0;
}

标签：游戏,minn,min,int,res,minid,SDOI2016,P4069,dis
来源： https://www.cnblogs.com/nofind/p/11991893.html