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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\n",ansp,anscost); } return 0; }
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原文地址:http://www.cnblogs.com/maijing/p/4657209.html