[模板] 虚树 && bzoj2286-[Sdoi2011]消耗战

简介

虚树可以解决一些关于树上一部分节点的问题. 对于一棵树 (T) 的一个子集 (S), 可以在 (O(|S| log |S|)) 的时间复杂度内求出 (S) 的虚树.

虚树包括根节点, 所有询问点和所有询问点之间的 (lca).

代码

//store the tree
struct tg{
	struct te{int t,pr,v;}edge[nsz*2];
	int hd[nsz],pe=1;
	void adde(int f,int t,int v){edge[++pe]=(te){t,hd[f],v};hd[f]=pe;}
	void adddb(int f,int t,int v){adde(f,t,v);adde(t,f,v);}
#define forg(g,p,i,v) for(int i=g.hd[p],v=g.edge[i].t;i;i=g.edge[i].pr,v=g.edge[i].t)
	void clear(){//only g2
		pe=1;
//		rep(i,1,pu)hd[usedp[i]]=0;
	}
}g1,g2;

//get lca (with sparse table)
namespace nlca{
	void dfs(int u,int fa){
	    eul[++pe]=u,vis[u]=pe,d[u]=d[fa]+1;
	    forg(g1,u,i,v){
	        if(v==fa)continue;
	        dfs(v,u);
	        eul[++pe]=u;
	    }
	}
	int dmin(int a,int b){return d[a]<=d[b]?a:b;}
	void rmq(){
	    rep(i,1,pe)stt[i][0]=eul[i];
	    rep(j,1,l2n[pe]){
	        rep(i,1,pe+1-(1<<j)){
	            stt[i][j]=dmin(stt[i][j-1],stt[i+(1<<(j-1))][j-1]);
	        }
	    }
	}
	int stqu(int a,int b){
	    int l=l2n[b-a+1];
	    return dmin(stt[a][l],stt[b-(1<<l)+1][l]);
	}
	void eulinit(){
	    int l=0;
	    rep(i,1,n*3){
	        if(i==(1<<(l+1)))++l;
	        l2n[i]=l;
	    }
	    dfs(1,0);
	    rmq();
	}
	int lca(int a,int b){
	    int x=vis[a],y=vis[b];
	    if(x>y)swap(x,y);
	    return stqu(x,y);
	}
}

// 求虚树
// line[1...k]: 用到的点
bool cmp(int a,int b){return vis[a]<vis[b];}
int stk[nsz],top=0;
void build(){
	g2.clear(),top=0;
	sort(line+1,line+k+1,cmp);
	top=0,stk[++top]=1;
	rep(i,1,k){
		int l=nlca::lca(line[i],stk[top]);
		while(top>1&&vis[stk[top-1]]>=vis[l])g2.adde(stk[top-1],stk[top],1),--top;
		if(l!=stk[top])g2.adde(l,stk[top],1),stk[top]=l;
		stk[++top]=line[i];
	}
	while(top>1)g2.adde(stk[top-1],stk[top],1),--top;
}

//dfs 过程
void sol(int p){
	forg(g2,p,i,v){
		sol(v);
		//do something...
	}
	g2.hd[p]=0; //清空虚树
}

例题: BZOJ2286 [Sdoi2011]消耗战

建立虚树之后dp即可.

注意输入的节点必须断掉, 但lca节点可断可不断. 可以标记输入的节点.

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<set>
#include<map>
using namespace std;
#define rep(i,l,r) for(register int i=(l);i<=(r);++i)
#define repdo(i,l,r) for(register int i=(l);i>=(r);--i)
#define il inline
typedef double db;
typedef long long ll;

//---------------------------------------
const int nsz=250050;
const ll ninf=1e18;

int n,m,k,line[nsz];
int used[nsz];

struct tg{
	struct te{int t,pr,v;}edge[nsz*2];
	int hd[nsz],pe=1;
	void adde(int f,int t,int v){edge[++pe]=(te){t,hd[f],v};hd[f]=pe;}
	void adddb(int f,int t,int v){adde(f,t,v);adde(t,f,v);}
#define forg(g,p,i,v) for(int i=g.hd[p],v=g.edge[i].t;i;i=g.edge[i].pr,v=g.edge[i].t)
	void clear(){//only g2
		pe=1;
//		rep(i,1,pu)hd[usedp[i]]=0;
	}
}g1,g2;

ll mind[nsz]{0,ninf};

int l2n[nsz*3+50];
int eul[nsz*3],pe=0,vis[nsz],d[nsz];
int stt[nsz*3][21];

namespace nlca{
	void dfs(int u,int fa){
	    eul[++pe]=u,vis[u]=pe,d[u]=d[fa]+1;
	    forg(g1,u,i,v){
	        if(v==fa)continue;
			mind[v]=min(mind[u],(ll)g1.edge[i].v);
	        dfs(v,u);
	        eul[++pe]=u;
	    }
	}
	int dmin(int a,int b){return d[a]<=d[b]?a:b;}
	void rmq(){
	    rep(i,1,pe)stt[i][0]=eul[i];
	    rep(j,1,l2n[pe]){
	        rep(i,1,pe+1-(1<<j)){
	            stt[i][j]=dmin(stt[i][j-1],stt[i+(1<<(j-1))][j-1]);
	        }
	    }
	}
	int stqu(int a,int b){
	    int l=l2n[b-a+1];
	    return dmin(stt[a][l],stt[b-(1<<l)+1][l]);
	}
	void eulinit(){
	    int l=0;
	    rep(i,1,n*3){
	        if(i==(1<<(l+1)))++l;
	        l2n[i]=l;
	    }
	    dfs(1,0);
	    rmq();
	}
	int lca(int a,int b){
	    int x=vis[a],y=vis[b];
	    if(x>y)swap(x,y);
	    return stqu(x,y);
	}
}

bool cmp(int a,int b){return vis[a]<vis[b];}
int stk[nsz],top=0;
void build(){
	g2.clear(),top=0;
	sort(line+1,line+k+1,cmp);
	top=0,stk[++top]=1;
	rep(i,1,k){
		int l=nlca::lca(line[i],stk[top]);
		while(top>1&&vis[stk[top-1]]>=vis[l])g2.adde(stk[top-1],stk[top],1),--top;
		if(l!=stk[top])g2.adde(l,stk[top],1),stk[top]=l;
		stk[++top]=line[i];
	}
	while(top>1)g2.adde(stk[top-1],stk[top],1),--top;
}

ll dp[nsz];
void sol(int p){
	dp[p]=0;
	forg(g2,p,i,v){
		sol(v);
		dp[p]+=(used[v]?mind[v]:min(mind[v],dp[v]));
	}
	g2.hd[p]=0;
}

int main(){
	ios::sync_with_stdio(0),cin.tie(0);
	cin>>n;
	int a,b,c;
	rep(i,2,n){
		cin>>a>>b>>c;
		g1.adddb(a,b,c);
	}
	nlca::eulinit();
	cin>>m;
	rep(i,1,m){
		cin>>k;
		rep(j,1,k)cin>>line[j],used[line[j]]=1;
		pe=1;
		build();
		sol(1);
		cout<<dp[1]<<'
';
		rep(j,1,k)used[line[j]]=0;
	}
	return 0;
}