【bzoj1787】[Ahoi2008]Meet 紧急集合

1787: [Ahoi2008]Meet 紧急集合

Time Limit: 20 Sec  Memory Limit: 162 MB
Submit: 2466  Solved: 1117
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Description

Input

Output

Sample Input

6 4
1 2
2 3
2 4
4 5
5 6
4 5 6
6 3 1
2 4 4
6 6 6

Sample Output

5 2
2 5
4 1
6 0

 

 

【题解】

记住一个结论：三个点的最小距离点等于两两的lca中与其他两个lca不同的lca

然后就很水了。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cstdlib>
 #include<ctime>
 #include<cmath>
 #include<algorithm>
 using namespace std;
 #define MAXN 500010
 struct node{int y,next;}e[MAXN*];
 int n,m,len,ans,Link[MAXN],f[MAXN],deep[MAXN],anc[MAXN][];
 inline int read()
 {
     int x=,f=;  char ch=getchar();
     while(!isdigit(ch))  {if(ch=='-')  f=-;  ch=getchar();}
     while(isdigit(ch))  {x=x*+ch-'';  ch=getchar();}
     return x*f;
 }
 void insert(int x,int y)  {e[++len].next=Link[x];Link[x]=len;e[len].y=y;}
 void dfs(int x,int fa)
 {
     anc[x][]=f[x];
     for(int i=;i<=;i++)  anc[x][i]=anc[anc[x][i-]][i-];
     for(int i=Link[x];i;i=e[i].next)
         if(e[i].y!=fa)
         {
             deep[e[i].y]=deep[x]+;
             f[e[i].y]=x;
             dfs(e[i].y,x);
         }
 }
 int lca(int x,int y)
 {
     if(deep[x]<deep[y])  swap(x,y);
     for(int i=;i>=;i--)  if(deep[anc[x][i]]>=deep[y])  x=anc[x][i];
     if(x==y)  return x;
     for(int i=;i>=;i--)  if(anc[x][i]!=anc[y][i])  x=anc[x][i],y=anc[y][i];
     return f[x];
 }
 int cal(int a,int b,int c)  {if(a==b)return c;else if(a==c)return b;else return a;}
 int dis(int x,int y)  {int t=lca(x,y);return deep[x]+deep[y]-*deep[t];}
 int main()
 {
     //freopen("cin.in","r",stdin);
     //freopen("cout.out","w",stdout);
     n=read();  m=read();
     for(int i=;i<n;i++)  {int x=read(),y=read();  insert(x,y);  insert(y,x);}
     deep[]=;  dfs(,);
     for(int i=;i<=m;i++)
     {
         int x=read(),y=read(),z=read();
         int a=lca(x,y),b=lca(x,z),c=lca(y,z);
         int p=cal(a,b,c);
         ans=dis(x,p)+dis(y,p)+dis(z,p);
         printf("%d %d\n",p,ans);
     }
     return ;
 }