1.bzoj1787紧急集合

找一点使这一点到三个点有最短距离 枚举三个两两LCA即可

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#define inf 999999999
using namespace std;
int n,q,cnt;
int deep[500001],head[500001],fa[500001][20];
bool vis[500001];
struct data{
	int to,next;
}e[1000001];
void ins(int u,int v)
{	
	e[++cnt].to=v;
	e[cnt].next=head[u];
	head[u]=cnt;
}
void insert(int u,int v)
{
	ins(u,v);
	ins(v,u);
}
void dfs(int x)
{
    vis[x]=1;
    for(int i=1;i<=18;i++)
    {
        if(deep[x]<(1<<i))break;
        fa[x][i]=fa[fa[x][i-1]][i-1];
    }
    for(int i=head[x];i;i=e[i].next)
    {
        if(vis[e[i].to])continue;
        deep[e[i].to]=deep[x]+1;
        fa[e[i].to][0]=x;
        dfs(e[i].to);
    }
}
int lca(int x,int y)
{
    if(deep[x]<deep[y])swap(x,y);
    int t=deep[x]-deep[y];
    for(int i=0;i<=18;i++)
       if(t&(1<<i))x=fa[x][i];
    for(int i=18;i>=0;i--)
       if(fa[x][i]!=fa[y][i])
       {x=fa[x][i];y=fa[y][i];}
    if(x==y)return x;
    return fa[x][0];
}
int val(int x,int y,int z)
{
	int p1=lca(x,y),p2=lca(x,z),p3=lca(y,z);
	int ans=inf,tmp,t;
	int q1=lca(p1,z),q2=lca(p2,y),q3=lca(p3,x);
	tmp=deep[x]+deep[y]-deep[p1]+deep[z]-2*deep[q1];
	if(tmp<ans){ans=tmp;t=p1;}
	
	tmp=deep[x]+deep[z]-deep[p2]+deep[y]-2*deep[q2];
	if(tmp<ans){ans=tmp;t=p2;}
	
	tmp=deep[y]+deep[z]-deep[p3]+deep[x]-2*deep[q3];
	if(tmp<ans){ans=tmp;t=p3;}
	
	printf("%d %d\n",t,ans);
}
int main()
{
    scanf("%d%d",&n,&q);
    for(int i=1;i<n;i++)
    {
        int u,v;
        scanf("%d%d",&u,&v);
        insert(u,v);
    }
    dfs(1);
    for(int i=1;i<=q;i++)
    {
        int x,y,z;
        scanf("%d%d%d",&x,&y,&z);
        val(x,y,z);
    }
    return 0;
}

2.codevs1036商务旅行

基础题 就求LCA 即可 到一个城市改变now值

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#define N 60005
using namespace std;
int b[N],p[N],nt[N],deep[N],fa[N][25],vis[N];
int n,m;int num=0;
void insert(){
	scanf("%d",&n);
	int x,y;
	for(int i=1;i<=n-1;i++){
		scanf("%d%d",&x,&y);
		num++;
		b[num]=y;
		nt[num]=p[x];
		p[x]=num;

		num++;
		b[num]=x;
		nt[num]=p[y];
		p[y]=num;
	}
}
void dfs(int x){
    vis[x]=1;
	for(int i=1;i<=20;i++){
		if(deep[x]<(1<<i)) break;
		fa[x][i]=fa[fa[x][i-1]][i-1];
	}
	for(int i=p[x];i;i=nt[i]){
		int v=b[i];
		if(vis[v]) continue;
		fa[v][0]=x;
		deep[v]=deep[x]+1;
		dfs(v);
	}
}
int lca(int x,int y){
	if(deep[x]<deep[y]) swap(x,y);
	int t=deep[x]-deep[y];
	for(int i=0;i<=20;i++){
		if(t&(1<<i)) x=fa[x][i];
	}
	if(x==y) return x;
	for(int i=20;i>=0;i--){
		if(fa[x][i]!=fa[y][i])
		{
			x=fa[x][i];y=fa[y][i];
		}
	}
	return fa[x][0];
}
int main(){
	insert();
	dfs(1);
	scanf("%d",&m);
	int now=1,to,ans=0;
	for(int i=1;i<=m;i++){
		scanf("%d",&to);
		int t=lca(now,to);
		int dis=deep[now]+deep[to]-2*deep[t];
		//printf("t=%d %d to %d is %d\n",t,now,to,dis);
		ans+=dis;
		now=to;
	}
	printf("%d",ans);
	return 0;
}

3.codevs2370小机房的树

同样的基础题 比上面的还简单

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#define M 100005
#define N 100005
using namespace std;
int n,m;int num=0;
int b[M],w[M],nt[M],p[N];
int fa[M][20],d[M],val[N];
void insert(){
	scanf("%d",&n);
	for(int i=1;i<=n-1;i++){
		int x,y,z;
		scanf("%d%d%d",&x,&y,&z);
		num++;
		b[num]=y;
		w[num]=z;
		nt[num]=p[x];
		p[x]=num;

		num++;
		b[num]=x;
		w[num]=z;
		nt[num]=p[y];
		p[y]=num;
	}
}
void dfs(int x){
	for(int i=1;i<=20;i++){
		if(d[x]<(1<<i))break;
		fa[x][i]=fa[fa[x][i-1]][i-1];
	}
	for(int i=p[x];i>0;i=nt[i]){
		int v=b[i];
		if(v==fa[x][0]) continue;//无向边 不走已经走过的父节点 
		fa[v][0]=x;
		d[v]=d[x]+1;//求深度=父节点深度+1 
		val[v]=val[x]+w[i];//求与根结点距离 和深度类似
        dfs(v);
	}
}
int lca(int x,int y){//求LCA 
	int h;
	if(d[x]<d[y]) swap(x,y);
	for(h=0;(1<<h)<=d[x];h++);
	h--;
	for(int i=h;i>=0;i--){
		if(d[x]-(1<<i)>=d[y])
			x=fa[x][i];
	}
	if(x==y) return x;
	for(int i=h;i>=0;i--){
		if(fa[x][i]!=fa[y][i]){
			x=fa[x][i];
			y=fa[y][i];
		}
	}
	return fa[x][0]; 
}	 
int main(){
	insert();
	dfs(0);
	scanf("%d",&m);
	int x,y;
	for(int i=1;i<=m;i++){
		scanf("%d%d",&x,&y);
		printf("%d\n",val[x]+val[y]-2*val[lca(x,y)]);
	}
	
	return 0;
}

4.vijos1460拉力赛

记录深度数组和求LCA即可

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#define N 10005
#define ll long long
using namespace std;
int n,m;
int b[N],p[N],nt[N],fa[N][21],d[N];
ll w[N];
ll val[N];
ll cnt,ans;
void insert(){
	scanf("%d%d",&n,&m);
	int x,y,z;
	for(int i=1;i<=n-1;i++){
		scanf("%d%d%d",&x,&y,&z);
		b[i]=y;
		w[i]=z;
		nt[i]=p[x];
		p[x]=i;
	}
}
void dfs(int x){
	for(int i=1;i<=20;i++){
		if(d[x]<(1<<i)) break;
		fa[x][i]=fa[fa[x][i-1]][i-1];
	}
	for(int i=p[x];i;i=nt[i]){
		int v=b[i];
		if(v==fa[x][0]) continue;
		fa[v][0]=x;
		d[v]=d[x]+1;
		val[v]=val[x]+w[i];
		dfs(v);
	}
}
bool pdlca(int x,int y){
	int t=d[x]-d[y];
	for(int i=0;i<=20;i++)
		if(t&(1<<i)) x=fa[x][i];
	if(x==y) return 1;
	else return 0;
}
int main(){
	insert();
	dfs(1);
	int u,v;
	for(int i=1;i<=m;i++){
		scanf("%d%d",&u,&v);
		if(d[v]<=d[u]) continue;
		if(!pdlca(v,u)) continue;
		cnt++;
		ans+=val[v]-val[u];
	}
	printf("%lld\n",cnt);
	printf("%lld",ans);
	return 0;
}

5.zoj3195 design the city

求树上三个点到某点 

有最小距离和输出这个距离和方法是 

会求两点最小距离 就会求三点距离=三个两两之间的最短距离相加再除2 

(不明觉厉。。。)

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#define N 100005
using namespace std;
int b[N],p[N],nt[N],fa[N][30],val[N],w[N],deep[N];
int n,q,num=0;
void insert(){
    num=0;
	for(int i=1;i<=n-1;i++)
	{
		int x,y,z;
		scanf("%d%d%d",&x,&y,&z);
		num++;
		b[num]=y;
		w[num]=z;
		nt[num]=p[x];
		p[x]=num;
		num++;
		b[num]=x;
		w[num]=z;
		nt[num]=p[y];
		p[y]=num;
	}
}
void dfs(int x,int h){
	deep[x]=h;
	for(int i=1;i<=26;i++){
		if(deep[x]<(1<<i)) break;
		fa[x][i]=fa[fa[x][i-1]][i-1];
	}
	for(int i=p[x];i;i=nt[i]){
		int v=b[i];
		if(v==fa[x][0]) continue;
		fa[v][0]=x;
		val[v]=val[x]+w[i];
		dfs(v,h+1);
	}
}
int lca(int x,int y){
	if(deep[x]<deep[y]) swap(x,y);
	int t=deep[x]-deep[y];
	for(int i=0;i<=26;i++)
		if(t&(1<<i)) x=fa[x][i];
	if(x==y) return x;
	for(int i=26;i>=0;i--){
		if(fa[x][i]!=fa[y][i])
		{
			x=fa[x][i];y=fa[y][i];
		}
	}
	return fa[x][0];
}
void clear(){
	memset(fa,0,sizeof(fa));
	memset(p,0,sizeof(p));
	memset(val,0,sizeof(val));
}
int query(int x,int y){
	int t=lca(x,y);
	return val[x]+val[y]-2*val[t];
}
int deal(int x,int y,int z){
	int ans=0;
	ans=query(x,y)+query(y,z)+query(x,z);
	ans=ans>>1;
	return ans;
}
int main(){
	int x,y,z;
	int tot=0;
	while(scanf("%d",&n)!=EOF){
		if(tot++)  {printf("\n");}//如果最后多了一个空行就PE了  
		clear();
		insert();
		dfs(0,0);
		scanf("%d",&q);
		for(int i=1;i<=q;i++){
			scanf("%d%d%d",&x,&y,&z);
			printf("%d\n",deal(x,y,z));
		}
	}
	return 0;
}