【BZOJ4999】This Problem Is Too Simple!(线段树)

【BZOJ4999】This Problem Is Too Simple!(线段树)

题面

BZOJ

题解

对于每个值,维护一棵线段树就好啦

动态开点,否则空间开不下

剩下的就是很简单的问题啦

当然了,对于数值要离散化

没必要离线吧,在线用\(map\)维护就行了

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
#include<queue>
using namespace std;
#define ll long long
#define RG register
#define MAX 120000
#define lson (t[x].ls)
#define rson (t[x].rs)
inline int read()
{
    RG int x=0,t=1;RG char ch=getchar();
    while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
    if(ch=='-')t=-1,ch=getchar();
    while(ch<='9'&&ch>='0')x=x*10+ch-48,ch=getchar();
    return x*t;
}
struct Line{int v,next;}e[MAX<<1];
int h[MAX],cnt=1;
inline void Add(int u,int v){e[cnt]=(Line){v,h[u]};h[u]=cnt++;}
map<int,int> M;
int tim,dfn[MAX],hson[MAX],size[MAX],fa[MAX],top[MAX],C[MAX],dep[MAX];
void dfs1(int u,int ff)
{
    fa[u]=ff;size[u]=1;dep[u]=dep[ff]+1;
    for(int i=h[u];i;i=e[i].next)
    {
        int v=e[i].v;
        if(v==ff)continue;
        dfs1(v,u);
        if(size[v]>size[hson[u]])hson[u]=v;
        size[u]+=size[v];
    }
}
void dfs2(int u,int tp)
{
    top[u]=tp;dfn[u]=++tim;
    if(hson[u])dfs2(hson[u],tp);
    for(int i=h[u];i;i=e[i].next)
        if(e[i].v!=fa[u]&&e[i].v!=hson[u])
            dfs2(e[i].v,e[i].v);
}
struct Node{int ls,rs,v;}t[MAX<<7];
int tot,sum,rt[MAX<<2];
int n,Q;
void Modify(int &x,int l,int r,int p,int w)
{
    if(!x)x=++tot;t[x].v+=w;
    if(l==r)return;
    int mid=(l+r)>>1;
    if(p<=mid)Modify(lson,l,mid,p,w);
    else Modify(rson,mid+1,r,p,w);
}
int Query(int x,int l,int r,int L,int R)
{
    if((!t[x].v)||(!x))return 0;
    if(L<=l&&r<=R)return t[x].v;
    int mid=(l+r)>>1,ret=0;
    if(L<=mid)ret+=Query(lson,l,mid,L,R);
    if(R>mid)ret+=Query(rson,mid+1,r,L,R);
    return ret;
}
int Route(int u,int v,int x)
{
    int ret=0;
    while(top[u]!=top[v])
    {
        if(dep[top[u]]<dep[top[v]])swap(u,v);
        ret+=Query(rt[x],1,n,dfn[top[u]],dfn[u]);
        u=fa[top[u]];
    }
    if(dep[u]<dep[v])swap(u,v);
    ret+=Query(rt[x],1,n,dfn[v],dfn[u]);
    return ret;
}
int main()
{
    n=read();Q=read();
    for(int i=1;i<=n;++i)C[i]=read();
    for(int i=1;i<n;++i)
    {
        int u=read(),v=read();
        Add(u,v);Add(v,u);
    }
    dfs1(1,0);dfs2(1,1);
    for(int i=1;i<=n;++i)
    {
        if(!M[C[i]])M[C[i]]=++sum;
        Modify(rt[M[C[i]]],1,n,dfn[i],1);
    }
    char ch[5];
    while(Q--)
    {
        scanf("%s",ch);
        if(ch[0]=='C')
        {
            int u=read(),x=read();
            Modify(rt[M[C[u]]],1,n,dfn[u],-1);
            if(!M[x])M[x]=++sum;
            Modify(rt[M[x]],1,n,dfn[u],1);
            C[u]=x;
        }
        else
        {
            int u=read(),v=read(),x=read();
            if(!M[x])puts("0");
            else printf("%d\n",Route(u,v,M[x]));
        }
    }
    return 0;
}
posted @ 2018-03-13 15:32  小蒟蒻yyb  阅读(...)  评论(...编辑  收藏