【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*2];
int n,m,len,ans,Link[MAXN],f[MAXN],deep[MAXN],anc[MAXN][25];
inline int read()
{
    int x=0,f=1;  char ch=getchar();
    while(!isdigit(ch))  {if(ch=='-')  f=-1;  ch=getchar();}
    while(isdigit(ch))  {x=x*10+ch-'0';  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][0]=f[x];
    for(int i=1;i<=20;i++)  anc[x][i]=anc[anc[x][i-1]][i-1];
    for(int i=Link[x];i;i=e[i].next)
        if(e[i].y!=fa)
        {
            deep[e[i].y]=deep[x]+1;
            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=20;i>=0;i--)  if(deep[anc[x][i]]>=deep[y])  x=anc[x][i];
    if(x==y)  return x;
    for(int i=20;i>=0;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]-2*deep[t];}
int main()
{
    //freopen("cin.in","r",stdin);
    //freopen("cout.out","w",stdout);
    n=read();  m=read();
    for(int i=1;i<n;i++)  {int x=read(),y=read();  insert(x,y);  insert(y,x);}
    deep[1]=1;  dfs(1,0);
    for(int i=1;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 0;
}