• bzoj1787


    题目:http://www.lydsy.com/JudgeOnline/problem.php?id=1787

    刚开始好像很难的样子

    画了一下,发现好像只有3种情况,且最终的P点一定是两两LCA中的一个。

    裸的LCA

    怪不得数据那么大

    #include<cstdio>
    #include<cstdlib>
    #include<iostream>
    #include<fstream>
    #include<algorithm>
    #include<cstring>
    #include<string>
    #include<cmath>
    #include<queue>
    #include<stack>
    #include<map>
    #include<utility>
    #include<set>
    #include<bitset>
    #include<vector>
    #include<functional>
    #include<deque>
    #include<cctype>
    #include<climits>
    #include<complex>
    //#include<bits/stdc++.h>适用于CF,UOJ,但不适用于poj
     
    using namespace std;
    
    typedef long long LL;
    typedef double DB;
    typedef pair<int,int> PII;
    typedef complex<DB> CP;
    
    #define mmst(a,v) memset(a,v,sizeof(a))
    #define mmcy(a,b) memcpy(a,b,sizeof(a))
    #define re(i,a,b)  for(i=a;i<=b;i++)
    #define red(i,a,b) for(i=a;i>=b;i--)
    #define fi first
    #define se second
    #define m_p(a,b) make_pair(a,b)
    #define SF scanf
    #define PF printf
    #define two(k) (1<<(k))
    
    template<class T>inline T sqr(T x){return x*x;}
    template<class T>inline void upmin(T &t,T tmp){if(t>tmp)t=tmp;}
    template<class T>inline void upmax(T &t,T tmp){if(t<tmp)t=tmp;}
    
    const DB EPS=1e-9;
    inline int sgn(DB x){if(abs(x)<EPS)return 0;return(x>0)?1:-1;}
    const DB Pi=acos(-1.0);
    
    inline int gint()
      {
            int res=0;bool neg=0;char z;
            for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());
            if(z==EOF)return 0;
            if(z=='-'){neg=1;z=getchar();}
            for(;z!=EOF && isdigit(z);res=res*10+z-'0',z=getchar());
            return (neg)?-res:res; 
        }
    inline LL gll()
      {
          LL res=0;bool neg=0;char z;
            for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());
            if(z==EOF)return 0;
            if(z=='-'){neg=1;z=getchar();}
            for(;z!=EOF && isdigit(z);res=res*10+z-'0',z=getchar());
            return (neg)?-res:res; 
        }
    
    const int maxN=500000;
    const int maxM=500000;
    
    int N,M;
    int now,first[maxN+100];
    struct Tedge{int v,next;}edge[2*maxN+100];
    int fa[maxN+100],dep[maxN+100],jump[maxN+100][40];
    
    inline void addedge(int u,int v)
      {
          now++;
          edge[now].v=v;
          edge[now].next=first[u];
          first[u]=now;
      }
    
    int head,tail,que[maxN+100];
    inline void BFS()
      {
          int i,j;
          dep[que[head=tail=1]=1]=1;
          fa[1]=1;
          while(head<=tail)
            {
                int u=que[head++],v;
                for(i=first[u],v=edge[i].v;i!=-1;i=edge[i].next,v=edge[i].v)if(!dep[v]) dep[que[++tail]=v]=dep[u]+1,fa[v]=u;
            }
          re(j,1,tail)
            {
                int u=que[j];
                jump[u][0]=fa[u];
                re(i,1,30)jump[u][i]=jump[jump[u][i-1]][i-1];
            }
      }
    
    inline void swim(int &x,int H){for(int i=0;H!=0;H>>=1,i++)if(H&1)x=jump[x][i];}
    inline int lca(int x,int y)
      {
          if(dep[x]<dep[y]) swap(x,y);
          swim(x,dep[x]-dep[y]);
          if(x==y)return x;
          int i;red(i,30,0)if(jump[x][i]!=jump[y][i])x=jump[x][i],y=jump[y][i];
          return jump[x][0];
      }
    
    inline int dist(int a,int b){return dep[a]+dep[b]-2*dep[lca(a,b)];}
    
    int main()
      {
          freopen("bzoj1787.in","r",stdin);
          freopen("bzoj1787.out","w",stdout);
          int i;
          N=gint();M=gint();
          mmst(first,-1);now=-1;
          re(i,1,N-1){int x=gint(),y=gint();addedge(x,y);addedge(y,x);}
          BFS();
          while(M--)
            {
                int a=gint(),b=gint(),c=gint(),p,cost,ansp,anscost;
                ansp=lca(a,b);anscost=dist(a,ansp)+dist(b,ansp)+dist(c,ansp);
                p=lca(a,c);cost=dist(a,p)+dist(b,p)+dist(c,p);if(cost<anscost){anscost=cost;ansp=p;}
                p=lca(b,c);cost=dist(a,p)+dist(b,p)+dist(c,p);if(cost<anscost){anscost=cost;ansp=p;}
                PF("%d %d
    ",ansp,anscost);
            }
          return 0;
      }
                
    View Code
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  • 原文地址:https://www.cnblogs.com/maijing/p/4657209.html
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