BZOJ JLOI2014 松鼠的新家

来源：互联网发布：mac代购编辑：程序博客网时间：2024/04/24 17:03

我想说的是，MD，我的线段树就长得这么难看?

传送门
这是一道树链剖分的裸题
给你一棵树，按照给定的顺序，每次将两点的简单路径上的点点权+1，最后询问各点的点权（只在最后询问一次）

由于只询问一次，所以可以树上差分，每次将两个点的差分值++，LCA的差分值–，fa[LCA]的差分–，最够从底层上传标记就行了
差分数组版本

#include <stdio.h>#include <string.h>#include <algorithm>#include <iostream>using namespace std;#define N 300000+5#define M 600000+5int head[N],size[N];int top[N],fa[N],depth[N];int cha[N];struct graph{    int next,to;    graph () {}    graph (int _next,int _to)    :next(_next),to(_to){}}edge[M];inline void add(int x,int y){    static int cnt = 0;    edge[++cnt]=graph(head[x],y);    head[x]=cnt;    edge[++cnt]=graph(head[y],x);    head[y]=cnt;}void DFS1(int x){    size[x]=1;    for(int i=head[x];i;i=edge[i].next)        if(edge[i].to!=fa[x])        {            depth[edge[i].to]=depth[x]+1;            fa[edge[i].to]=x;            DFS1(edge[i].to);            size[x]+=size[edge[i].to];        }}void DFS2(int x,int chain){    top[x]=chain;    int k =0;    for(int i=head[x];i;i=edge[i].next)        if(edge[i].to!=fa[x]&&size[k]<size[edge[i].to])            k=edge[i].to;        if(!k)return;    DFS2(k,chain);    for(int i=head[x];i;i=edge[i].next)        if(edge[i].to!=fa[x]&&edge[i].to!=k)            DFS2(edge[i].to,edge[i].to);}int lca(int x,int y){    while(top[x]!=top[y])    {        if(depth[top[x]]<depth[top[y]])swap(x,y);        x=fa[top[x]];    }    return depth[x]<depth[y]?x:y;}char getc(){    static const int LEN = 1<<15;    static char buf[LEN],*S=buf,*T=buf;    if(S == T)    {        T = (S=buf)+fread(buf,1,LEN,stdin);        if(S == T)return EOF;    }    return *S++;}int read(){    static char ch;    static int D;    while(!isdigit(ch=getc()));    for(D=ch-'0'; isdigit(ch=getc());)        D=(D<<3)+(D<<1)+(ch-'0');    return D;}void Getans(int x){    for(int i=head[x];i;i=edge[i].next)        if(edge[i].to!=fa[x])        {            Getans(edge[i].to);            cha[x]+=cha[edge[i].to];        }}int q[N];int main(){    int n=read();    for(int i=1;i<=n;++i)        q[i]=read();    for(int i=1;i<n;++i)    {        int x=read(),y=read();        add(x,y);    }    DFS1(q[1]);    DFS2(q[1],q[1]);    for(int i=1;i<n;++i)    {        int t = lca(q[i],q[i+1]);        cha[q[i]]++;cha[q[i+1]]++;        cha[t]--;cha[fa[t]]--;    }    Getans(q[1]);    for(int i=2;i<=n;++i)cha[q[i]]--;    for(int i=1;i<=n;++i)printf("%d\n",cha[i]);}

线段树区间修改版本，我好像是调大了系统栈

#include <stdio.h>#include <string.h>#include <algorithm>#include <iostream>#define N 300000+5#define M 1200000+5#define stack_size (20001000) using namespace std;int stack[stack_size],bak; int head[N],size[N],fa[N],top[N],w[N],depth[N],n;int sz;struct graph{    int next,to;    graph () {}    graph (int _next,int _to)    :next(_next),to(_to){}}edge[M];inline void add(int x,int y){    static int cnt = 0;    edge[++cnt]=graph(head[x],y);    head[x]=cnt;    edge[++cnt]=graph(head[y],x);    head[y]=cnt;}void DFS1(int x){    size[x]=1;    for(int i=head[x];i;i=edge[i].next)        if(edge[i].to!=fa[x])        {            depth[edge[i].to]=depth[x]+1;            fa[edge[i].to]=x;            DFS1(edge[i].to);            size[x]+=size[edge[i].to];        }}void DFS2(int x,int chain){    w[x]=++sz;    top[x]=chain;    int k=0;    for(int i=head[x];i;i=edge[i].next)        if(edge[i].to!=fa[x]&&size[k]<size[edge[i].to])            k=edge[i].to;    if(!k)return;    DFS2(k,chain);    for(int i=head[x];i;i=edge[i].next)        if(edge[i].to!=fa[x]&&edge[i].to!=k)            DFS2(edge[i].to,edge[i].to);}void CallDFS()    {        __asm__ __volatile__        (            "mov %%esp,%0\n"          "mov %1,%%esp\n"          :"=g"(bak)            :"g"(stack+stack_size-1)            :        );        DFS1(1);    DFS2(1,1);    __asm__ __volatile__        (            "mov %0,%%esp\n"          :            :"g"(bak)            :        );    }   struct seg{    int l,r,mn,lazy;}tr[M];void build(int k,int l,int r){    tr[k].l=l,tr[k].r=r;    if(l==r)return;    int mid=(l+r)>>1;    build(k<<1,l,mid);    build(k<<1|1,mid+1,r);}void down(int k){    int tmp=tr[k].lazy;    tr[k].lazy=0;    if(tr[k].l==tr[k].r||!tmp)        return;    tr[k<<1].lazy+=tmp;    tr[k<<1|1].lazy+=tmp;    tr[k<<1].mn+=(tr[k<<1].r-tr[k<<1].l+1)*tmp;    tr[k<<1|1].mn+=(tr[k<<1|1].r-tr[k<<1|1].l+1)*tmp;}inline void updata(int k){    tr[k].mn=tr[k<<1].mn+tr[k<<1|1].mn;}void change(int k,int l,int r){    down(k);    if(l==tr[k].l&&r==tr[k].r)    {        tr[k].lazy++;        tr[k].mn+=(r-l+1);        return;    }    int mid=(tr[k].l+tr[k].r)>>1;    if(r<=mid)        change(k<<1,l,r);    else if(l>mid)        change(k<<1|1,l,r);    else    {        change(k<<1,l,mid);        change(k<<1|1,mid+1,r);    }    updata(k);}int map[N];void solvechange(int x,int y){    while(top[x]!=top[y])    {        if(depth[top[x]]<depth[top[y]])            swap(x,y);    //  printf("change %d %d  \n",top[x],x);        change(1,w[top[x]],w[x]);        x=fa[top[x]];    }    if(w[x]>w[y])        swap(x,y);    change(1,w[x],w[y]);}int ask(int k,int pos){    down(k);    if(tr[k].l==pos && tr[k].r==pos)        return tr[k].mn;    int mid=(tr[k].l+tr[k].r)>>1;    if(pos<=mid)        return ask(k<<1,pos);    else        return ask(k<<1|1,pos);}int ans[N];void Getans(int k){    if(tr[k].l==tr[k].r)    {        ans[tr[k].l]=tr[k].mn;        return;    }    Getans(k<<1);    Getans(k<<1|1);}char getc(){    static const int LEN = 1<<15;    static char buf[LEN],*S=buf,*T=buf;    if(S == T)    {        T = (S=buf)+fread(buf,1,LEN,stdin);        if(S == T)return EOF;    }    return *S++;}int read(){    static char ch;    static int D;    while(!isdigit(ch=getc()));    for(D=ch-'0'; isdigit(ch=getc());)        D=(D<<3)+(D<<1)+(ch-'0');    return D;}void print( int k ){    int num = 0;    char ch[17];    while(k>0) ch[++num] = k % 10, k /= 10;    while(num)         putchar( ch[num--]+48 );    putchar(10);}void check(){    puts("checking ****************************");    for(int i=1;i<=n;++i)        printf("%d %d\n",i,ask(1,w[i]));    puts("checking ****************************");}int main(){    cin>>n;    for(int i=1;i<=n;++i)        map[i]=read();    for(int i=1;i<n;++i)    {        int x,y;        x=read(),y=read();        add(x,y);    }    build(1,1,n);    CallDFS();    for( int i=1;i<n;++i)        solvechange(map[i],map[i+1]);    for(int i=1;i<=n;++i)    {        if(i!=map[1])            printf("%d\n",ask(1,w[i])-1);        else            printf("%d\n",ask(1,w[i]));    }}

这里写图片描述

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