• 【hdu6334】【2018Multi-University-Training Contest04】Problem C. Problems on a Tree


    维护1边的联通块和2边的联通块,合并的时候直接启发式合并。

    cdqz的大爷好强啊。

    #include<bits/stdc++.h>
    #define lson (o<<1)
    #define rson (o<<1|1)
    #define fi first
    #define sc second
    #define dbg(x) cout<<#x<<" = "<<(x)<<endl;
    typedef long long ll;
    typedef unsigned int uint;
    typedef unsigned long long ull;
    using namespace std;
    const double pi=acos(-1);
    const double eps=1e-6;
    inline int lowbit(int x){return x&(-x);}
    inline int read(){
        int f=1,x=0;char ch;
        do{ch=getchar();if(ch=='-')f=-1;}while(ch<'0'||ch>'9');
        do{x=x*10+ch-'0';ch=getchar();}while(ch>='0'&&ch<='9');
        return f*x;
    }
    template<typename T> inline T max(T x,T y,T z){return max(max(x,y),z);}
    template<typename T> inline T min(T x,T y,T z){return min(min(x,y),z);}
    template<typename T> inline T sqr(T x){return x*x;}
    template<typename T> inline void checkmax(T &x,T y){x=max(x,y);}
    template<typename T> inline void checkmin(T &x,T y){x=min(x,y);}
    template<typename T> inline void read(T &x){
    x=0;T f=1;char ch;do{ch=getchar();if(ch=='-')f=-1;}while(ch<'0'||ch>'9');
    do x=x*10+ch-'0',ch=getchar();while(ch<='9'&&ch>='0');x*=f;
    }
    template<typename A,typename B,typename C> inline A fpow(A x,B p,C yql){
        A ans=1;
        for(;p;p>>=1,x=1LL*x*x%yql)if(p&1)ans=1LL*x*ans%yql;
        return ans;
    }
    struct FastIO{
        static const int S=1310720;
        int wpos;char wbuf[S];
        FastIO():wpos(0) {}
        inline int xchar(){
            static char buf[S];
            static int len=0,pos=0;
            if(pos==len)pos=0,len=fread(buf,1,S,stdin);
            if(pos==len)return -1;
            return buf[pos++];
        }
        inline int read(){
            int c=xchar(),x=0;
            while(c<=32&&~c)c=xchar();
            if(c==-1)return -1;
            for(;'0'<=c&&c<='9';c=xchar())x=x*10+c-'0';
            return x;
        }
    }io;
    //#define read io.read
    const int N=200010;
    int add[N],tag2[N],fa1[N],fa2[N],size1[N],size2[N],fa[N],d[N];
    int n,m;
    vector<int> G[N];
    inline int find(int *fa,int x){return x==fa[x]?x:fa[x]=find(fa,fa[x]);}
    inline void merge2(int x,int y){
        int xx=find(fa1,x),yy=find(fa1,y);
        if(d[xx]<d[yy])swap(xx,yy);
        add[yy]-=size2[xx];yy=find(fa2,yy);
        fa2[xx]=yy;size2[yy]+=size2[xx];
        add[find(fa1,fa[yy])]+=size2[xx];
    }
    inline void merge1(int x,int y){
        int xx=find(fa1,x),yy=find(fa1,y);
        if(d[xx]<d[yy])swap(xx,yy);
        fa1[xx]=yy;size1[yy]+=size1[xx];
        add[yy]+=add[xx];
    }
    inline void dfs(int u){
        d[u]=d[fa[u]]+1;
        add[u]=0;
        for(int i=0;i<G[u].size();i++){
            int v=G[u][i];
            if(v==fa[u])continue;
            fa[v]=u;dfs(v);add[u]++;
        }
    }
    vector<pair<int,int> >p1,p2;
    inline void clear(){
        memset(add,0,sizeof(add));
        for(int i=1;i<=n;i++){
            fa1[i]=i;fa2[i]=i;size1[i]=1;size2[i]=1;G[i].clear();
        }
        p1.clear();p2.clear();
    }
    inline void work(){
        for(int i=1;i<n;i++){
            int u=read(),v=read(),w=read();
            G[u].push_back(v);G[v].push_back(u);
            if(w==1){
                p1.push_back(make_pair(u,v));
                //merge2(u,v);merge1(u,v);
            }
            else if(w==2)p2.push_back(make_pair(u,v));
        }
        dfs(1);
        for(int i=0;i<p1.size();i++){
            pair<int,int> p=p1[i];
            merge2(p.fi,p.sc);merge1(p.fi,p.sc);
        }
        for(int i=0;i<p2.size();i++){
            pair<int,int> p=p2[i];
            merge2(p.fi,p.sc);
        }
        while(m--){
            int u=read(),v=read(),t=read(),s=read();
            u=find(fa1,u);v=find(fa1,v);
            if(u!=v){
                if(find(fa2,u)==find(fa2,v))merge1(u,v);
                else merge2(u,v);
            }
            s=find(fa1,s);t=find(fa1,t);int flag=0;
            if(find(fa2,s)==find(fa2,t))printf("1 "),flag=1;
            else{
                if(d[find(fa2,s)]>d[find(fa2,t)]){
                    if(find(fa1,fa[find(fa2,s)])==t)printf("1 "),flag=1;
                }
                else if((t==find(fa2,t)&&find(fa2,fa[t])==find(fa2,s)))printf("1 "),flag=1;
            }
            if(!flag)printf("0 ");
            printf("%d
    ",size2[find(fa2,t)]+add[t]+(t==find(fa2,t)?size2[find(fa2,fa[t])]:0));
        }
    }
    int main(){
        int T=read();
        while(T--){
            n=read();m=read();
            clear();
            work();
        }
    }
  • 相关阅读:
    Google Maps API 进级:通过XML文档加载Gpolyline或者Gpolygon
    Google Maps API 进级: GPolygon对象2
    Google Maps API 进级: GoogleMaps常用事件及应用思路2
    debian下安装JDK
    sudo 提示用户不在sudoers文件错误
    debian关闭开机自动启动时候的gui
    debian解决中文乱码,安装chinese font
    debian下intel 3945abg和T420无线网卡驱动
    debian修改系统时区
    wheezy安装oracle jdk的jdk之中文乱码处理
  • 原文地址:https://www.cnblogs.com/zcysky/p/9405009.html
Copyright © 2020-2023  润新知