题目地址:http://codeforces.com/problemset/problem/342/E
思路:dist[i]表示i到最近节点的距离。当有一定数量的红点未处理时,统一进行一次BFS,获得初步的dist值。对于每两个询问,用LCA求得最近距离,最后与dist[i]取最小值即为答案。
#include<cstdio>
#include<queue>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#define debu
using namespace std;
const int INF=1e9;
const int maxn=1e5+50;
struct Node
{
int v,w;
Node(int v=0,int w=0):v(v),w(w) {}
};
int n,m;
queue<int> q;
vector<int> red;
vector<Node> g[maxn];
int Dist[maxn],dist[maxn];
int anc[maxn][50];
int v[maxn],d[maxn],mm[maxn][50];
void dfs(int x,int dep)
{
v[x]=1,d[x]=dep;
for(int i=0; i<g[x].size(); i++)
{
int nt=g[x][i].v;
if(!v[nt])
{
anc[nt][0]=x;
mm[nt][0]=g[x][i].w;
int k=0;
while(anc[anc[nt][k]][k]!=0)
{
anc[nt][k+1]=anc[anc[nt][k]][k];
k++;
}
dfs(nt,dep+1);
}
}
}
int mindistanc(int x,int y)
{
int xx=x,yy=y;
if(d[x]<d[y]) swap(x,y);
int l=d[x]-d[y];
int k=0,ans=INF;
while(l!=0)
{
if(l&1) x=anc[x][k];
l>>=1;
k++;
}
k=0;
while(x!=y)
{
if((anc[x][k]!=anc[y][k])||(!k))
{
x=anc[x][k],y=anc[y][k];
k++;
}
else k--;
}
return Dist[xx]+Dist[yy]-2*Dist[x];
}
void init()
{
red.clear();
memset(v,0,sizeof(v));
memset(d,0,sizeof(d));
memset(anc,0,sizeof(anc));
memset(dist,0x3f3f3f3f,sizeof(dist));
for(int i=0; i<=n; i++) g[i].clear();
}
void solve()
{
memset(v,0,sizeof(v));
while(!q.empty()) q.pop();
for(int i=0; i<red.size(); i++)
{
v[red[i]]=1;
dist[red[i]]=0;
q.push(red[i]);
}
while(!q.empty())
{
int now=q.front();
q.pop(),v[now]=0;
for(int i=0; i<g[now].size(); i++)
{
int nt=g[now][i].v;
if(dist[nt]>dist[now]+1)
{
dist[nt]=dist[now]+1;
if(!v[nt])
{
v[nt]=1;
q.push(nt);
}
}
}
}
}
void rootdist()
{
memset(v,0,sizeof(v));
while(!q.empty()) q.pop();
memset(Dist,0x3f3f3f3f,sizeof(Dist));
q.push(1),v[1]=1,Dist[1]=0;
while(!q.empty())
{
int now=q.front();
q.pop(),v[now]=0;
for(int i=0; i<g[now].size(); i++)
{
int nt=g[now][i].v;
if(Dist[nt]>Dist[now]+1)
{
Dist[nt]=Dist[now]+1;
if(!v[nt])
{
v[nt]=1;
q.push(nt);
}
}
}
}
}
int main()
{
#ifdef debug
freopen("in.in","r",stdin);
#endif // debug
while(scanf("%d%d",&n,&m)!=EOF)
{
init();
red.push_back(1);
for(int i=1; i<=n-1; i++)
{
int x,y;
scanf("%d%d",&x,&y);
g[x].push_back(Node(y,1));
g[y].push_back(Node(x,1));
}
rootdist();
for(int i=1; i<=n; i++) if(!v[i]) dfs(i,0);
for(int i=1; i<=m; i++)
{
int t,p;
scanf("%d%d",&t,&p);
if(t==1)
{
red.push_back(p);
if(red.size()==floor(sqrt(m*1.0)))
{
solve();
red.clear();
}
}
else
{
int ans=dist[p];
for(int i=0; i<red.size(); i++)
{
//cout<<i<<" "<<p<<" "<<red[i]<<endl;
ans=min(ans,mindistanc(p,red[i]));
}
printf("%d\n",ans);
}
}
}
return 0;
}