树的统计 树链剖分

 

Code:

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long ll;
const int maxn=30000+5;
int head[maxn],nex[maxn*2],to[maxn*2];               
int dep[maxn],siz[maxn],son[maxn],p[maxn],top[maxn];
int A[maxn],maxv[maxn*4+3], val[maxn];  
int val2[maxn];                            
ll sumv[maxn*4+3];    
int _max;                
int cnt,n,cnt2;
void addedge(int u,int v)
{
	nex[++cnt]=head[u];
	head[u]=cnt,to[cnt]=v;
}
void dfs1(int u,int fa,int cur){
	p[u]=fa,dep[u]=cur,siz[u]=1;
    for(int i=head[u];i;i=nex[i])
	{
		int v=to[i];
		if(v!=fa)
		{
			dfs1(v,u,cur+1);
			siz[u]+=siz[v];
			if(son[u]==-1||siz[v]>siz[son[u]])son[u]=v;
		}
	}
}
void dfs2(int u,int tp)
{
	top[u]=tp,A[u]=++cnt2;
	if(son[u]>0)dfs2(son[u],tp);
	for(int i=head[u];i;i=nex[i])
	{
		int v=to[i];
		if(v!=p[u]&&v!=son[u])dfs2(v,v);
	}
}
void build_tree(int L,int R,int o,int arr[])
{
	if(L==R){
		sumv[o]=maxv[o]=arr[L];
		return;
	}
	int mid=(L+R)/2;
	build_tree(L,mid,o*2,arr);
	build_tree(mid+1,R,o*2+1,arr);
	sumv[o]=sumv[o*2]+sumv[o*2+1];
	maxv[o]=max(maxv[o*2],maxv[o*2+1]);
}
void update(int l,int k,int L,int R,int o)
{
	if(L==R)
	{
       sumv[o]=maxv[o]=k;
       return;
	}
	int mid=(L+R)/2;
	if(l<=mid)update(l,k,L,mid,o*2);
	else update(l,k,mid+1,R,o*2+1);
	sumv[o]=sumv[o*2]+sumv[o*2+1];
	maxv[o]=max(maxv[o*2],maxv[o*2+1]);
}
ll query(int l,int r,int L,int R,int o)
{
	if(l<=L&&r>=R)
	{
		_max=max(_max,maxv[o]);
		return sumv[o];
	}
	int mid=(L+R)/2;
	ll ret=0;
	if(l<=mid)ret+=query(l,r,L,mid,o*2);
	if(r>mid)ret+=query(l,r,mid+1,R,o*2+1);
	return ret;
}
ll lca(int x,int y)    
{
	_max=-300000;
	ll ret=0;
	while(top[x]!=top[y])
	{
         if(dep[top[x]]<dep[top[y]]){ret+=query(A[top[y]],A[y],1,n,1);y=p[top[y]];}
         else {ret+=query(A[top[x]],A[x],1,n,1);x=p[top[x]];}
	}
	if(dep[x]<dep[y])ret+=query(A[x],A[y],1,n,1);
	else ret+=query(A[y],A[x],1,n,1);
	return ret;
}
int main()
{
	memset(son,-1,sizeof(son));
	scanf("%d",&n);
	for(int i=1;i<n;++i){
		int a,b;
		scanf("%d%d",&a,&b);
		addedge(a,b);
		addedge(b,a);
	}
	for(int i=1;i<=n;++i)scanf("%d",&val[i]);
	dfs1(1,-1,1);                             //以一为根
    dfs2(1,1);
    for(int i=1;i<=n;++i)val2[A[i]]=val[i];
	build_tree(1,n,1,val2);                    //建树
	int T;
	scanf("%d",&T);
	while(T--)
	{
		char S[20];
		scanf("%s",S);
		if(S[0]=='C'){
			int u,t;
			scanf("%d%d",&u,&t);
			update(A[u],t,1,n,1);
		}
		if(S[1]=='M'){
            int x,y;
            scanf("%d%d",&x,&y);
            lca(x,y);
            printf("%d\n",_max);
		}
		if(S[1]=='S')
		{
			int x,y;
			scanf("%d%d",&x,&y);
			printf("%lld\n",lca(x,y));
		}
	}
	return 0;
}

  

posted @ 2018-10-16 16:55  EM-LGH  阅读(152)  评论(0编辑  收藏  举报