CodeForces

http://codeforces.com/problemset/problem/832/D

题意：给定一棵树，Misha和Grisha跳三个点a，b，c。两个人选其中两个为起点，剩下一个为终点出发。所有选择中要使两个人经过相同结点数量最多，输出最多的结点数。

题解：选定两个点a，b求最近公共祖先k，也就是LCA。这样就可以求出a，b之间的距离dis(a,b)=deep[a]+deep[b]-2*deep[k]。所以ans=(dis(a,b)+dis(a,c)-dis(b,c))/2+1(结点个数等于距离+1)。LCA直接套着模板，虽然倍增法思想是懂的，但是dfs那段还不太理解，需要深入学习。

代码：

#include<set>#include<map>#include<stack>#include<queue>#include<vector>#include<string>#include<bitset>#include<algorithm>#include<cstring>#include<cstdio>#include<cmath>#include<iomanip>#include<iostream>#define debug cout<<"aaa"<<endl#define d(a) cout<<a<<endl#define mem(a,b) memset(a,b,sizeof(a))#define LL long long#define lson l,mid,root<<1#define rson mid+1,r,root<<1|1#define MIN_INT (-2147483647-1)#define MAX_INT 2147483647#define MAX_LL 9223372036854775807i64#define MIN_LL (-9223372036854775807i64-1)using namespace std;const int N = 100000 + 5;const int mod = 1000000000 + 7;const double eps = 1e-8;int next[N*2],to[N*2],head[N*2],num,deep[N*2],fa[N*2][21];int n,m,rt,a,b,c,ff,ans;void add(int false_from,int false_to){next[++num]=head[false_from];to[num]=false_to;head[false_from]=num;}void dfs(int x){deep[x]=deep[fa[x][0]]+1;for(int i=0;fa[x][i];i++){fa[x][i+1]=fa[fa[x][i]][i];}for(int i=head[x];i;i=next[i]){if(!deep[to[i]]){fa[to[i]][0]=x;dfs(to[i]);}}}int lca(int x,int y){if(deep[x]>deep[y]){swap(x,y);}for(int i=20;i>=0;i--){if(deep[fa[y][i]]>=deep[x]){y=fa[y][i];}}if(x==y){return y;}for(int i=20;i>=0;i--){if(fa[y][i]!=fa[x][i]){y=fa[y][i];x=fa[x][i];}}return fa[x][0];}int dis(int u,int v){ff=lca(u,v);return deep[u]+deep[v]-2*deep[ff];}int main(){int lab,lac,lbc;scanf("%d%d",&n,&m);for(int i=2;i<=n;i++){scanf("%d",&a);add(a,i),add(i,a);}dfs(1);for(int i=1;i<=m;i++){ans=0;scanf("%d%d%d",&a,&b,&c);lab=dis(a,b);lac=dis(a,c);lbc=dis(c,b);ans=max((lab+lac-lbc)/2,ans);ans=max((lab+lbc-lac)/2,ans);ans=max((lac+lbc-lab)/2,ans);printf("%d\n",ans+1);}return 0;}