zoukankan      html  css  js  c++  java
  • 洛谷P4004 Hello world!(分块+并查集)

    传送门

    虽然洛谷数据水,然而咱最终还是没有卡过uoj上的毒瘤数据……

    神tm全uoj就3个人过了这题……

    首先,每个数最多被开根(6)次,开到(1)之后就别管它了,把它用并查集连到它父亲上

    它每次要走(k)步,我们可以对(k)分类讨论,如果(k)比较大就直接暴力跳,否则建(k)棵树,每次树剖+线段树维护

    然而我只会暴力跳了……

    //minamoto
    #include<bits/stdc++.h>
    #define R register
    #define ll long long
    #pragma GCC optimize(3)
    #define fp(i,a,b) for(R int i=a,I=b+1;i<I;++i)
    #define fd(i,a,b) for(R int i=a,I=b-1;i>I;--i)
    #define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)
    inline void swap(R int &x,R int &y){x^=y^=x^=y;}
    using namespace std;
    char buf[1<<21],*p1=buf,*p2=buf;
    inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}
    ll read(){
        R ll res,f=1;R char ch;
        while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);
        for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');
        return res*f;
    }
    char sr[1<<21],z[20];int C=-1,Z=0;
    inline void Ot(){fwrite(sr,1,C+1,stdout),C=-1;}
    void print(R ll x){
        if(C>1<<20)Ot();if(x<0)sr[++C]='-',x=-x;
        while(z[++Z]=x%10+48,x/=10);
        while(sr[++C]=z[Z],--Z);sr[++C]='
    ';
    }
    const int N=1e5+5,S=655;
    struct eg{int v,nx;}e[N<<1];int head[N],tot;
    inline void add(R int u,R int v){e[++tot]={v,head[u]},head[u]=tot;}
    ll a[N],ans;int ga[N],fa[N],fat[N][25],jp[N][S],dep[N],son[N],sz[N],top[N];
    int n,s,m,u,v,op,k;
    int find(R int x){return ga[x]==x?x:ga[x]=find(ga[x]);}
    void dfs1(int u,int f){
        sz[u]=1,dep[u]=dep[f]+1,fa[u]=fat[u][0]=jp[u][1]=f;
        jp[u][0]=u;
        fp(i,1,18)fat[u][i]=fat[fat[u][i-1]][i-1];
        fp(i,2,s)jp[u][i]=jp[f][i-1];
        go(u)if(v!=f){
            dfs1(v,u),sz[u]+=sz[v];
            if(sz[v]>sz[son[u]])son[u]=v;
        }
    }
    void dfs2(int u,int t){
        top[u]=t;if(!son[u])return;
        dfs2(son[u],t);
        go(u)if(!top[v])dfs2(v,v);
    }
    int LCA(R int u,R int v){
        while(top[u]!=top[v]){
            if(dep[top[u]]<dep[top[v]])swap(u,v);
            u=fa[top[u]];
        }return dep[u]<dep[v]?u:v;
    }
    struct query{
    	int md[N],len[N],top[N],son[N],hi[N];
    	vector<int>U[N],D[N];
    	void dfs1(int u,int fa){
    		md[u]=dep[u];
    		go(u)if(v!=fa){
    			dfs1(v,u);
    			if(md[v]>md[son[u]])son[u]=v,md[u]=md[v];
    		}
    	}
    	void dfs2(int u,int t){
    		top[u]=t,len[u]=md[u]-dep[t]+1;
    		if(!son[u])return;
    		dfs2(son[u],t);
    		go(u)if(!top[v])dfs2(v,v);
    	}
    	void init(){
    		dfs1(1,0),dfs2(1,1);
    		int now=0;
    		fp(i,1,n){
    			if(!(i>>now&1))++now;
    			hi[i]=now;
    		}
    		fp(i,1,n)if(i==top[i]){
    			for(R int j=1,u=i;j<=len[i]&&u;++j)u=fa[u],U[i].push_back(u);
    			for(R int j=1,u=i;j<=len[i]&&u;++j)u=son[u],D[i].push_back(u);
    		}
    	}
    	int Fa(int u,int k){
    		if(k>dep[u])return 0;if(!k)return u;
    		u=fat[u][hi[k]],k^=(1<<hi[k]);
    		if(!k)return u;
    		if(dep[u]-dep[top[u]]==k)return top[u];
    		if(k>dep[u]-dep[top[u]])return U[top[u]][k-dep[u]+dep[top[u]]-1];
    		return D[top[u]][dep[u]-dep[top[u]]-k-1];
    	}
    }qwq;
    void upd(R int u){
        if(a[u]==1)return;
        a[u]=sqrt(a[u]);
        if(a[u]==1)ga[u]=find(fa[u]);
    }
    int pos(int u,int v,int lca,int k){
        if(dep[v]-dep[lca]>=k)return qwq.Fa(v,k);
        k-=dep[v]-dep[lca];
        return qwq.Fa(u,dep[u]-dep[lca]-k);
    }
    int jump(int u,int k){
        if(k>s)return qwq.Fa(u,k);
        int v=find(fa[u]),d=(dep[u]-dep[v])%k;
        if(d)d=k-d;
        return jp[v][d];
    }
    void update(int u,int v,int k){
    	if(u==v)return upd(u);
        int lca=LCA(u,v),dis=dep[u]+dep[v]-(dep[lca]<<1);
        if(dis<=k)return upd(u),upd(v);
        if(dis%k)upd(v),v=pos(u,v,lca,dis%k),lca=LCA(u,v);
        while(dep[u]>=dep[lca])upd(u),u=jump(u,k);
        while(dep[v]>dep[lca])upd(v),v=jump(v,k);
    }
    void query(int u,int v,int k){
    	if(u==v)return ans=a[u],void();
        int lca=LCA(u,v),dis=dep[u]+dep[v]-(dep[lca]<<1);
        if(dis<=k)return ans=a[u]+a[v],void();
        if(dis%k)ans+=a[v],v=pos(u,v,lca,dis%k),lca=LCA(u,v);
        ans+=(dep[u]+dep[v]-(dep[lca]<<1))/k+1;
        while(dep[u]>=dep[lca])ans+=a[u]-1,u=jump(u,k);
        while(dep[v]>dep[lca])ans+=a[v]-1,v=jump(v,k);
    }
    int main(){
    //	freopen("testdata.in","r",stdin);
        n=read();
        fp(i,1,n)a[i]=read(),ga[i]=i;
        fp(i,1,n-1)u=read(),v=read(),add(u,v),add(v,u);
        m=read(),s=500;
        dfs1(1,0),dfs2(1,1),qwq.init();
        fp(i,1,n)if(a[i]==1)ga[i]=find(fa[i]);
        while(m--){
            op=read(),u=read(),v=read(),k=read();
            if(op==1){
            	ans=0,query(u,v,k),print(ans);
    		}else update(u,v,k);
        }
        return Ot(),0;
    }
    
  • 相关阅读:
    CF1359D Yet Another Yet Another Task
    【数据结构】fhq_treap
    AtCoder Beginner Contest 182 题解
    UVA11992 Fast Matrix Operations
    双指针例题
    python使用国内镜像库
    APP元素定位工具之——Weditor
    安卓ADB的常见命令的使用
    函数进阶之迭代器,递归
    函数基础之对象,嵌套,名称空间和作用域
  • 原文地址:https://www.cnblogs.com/bztMinamoto/p/10277912.html
Copyright © 2011-2022 走看看