HDU 5293(Tree chain problem-树链剖分)

Tree chain problem

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 940    Accepted Submission(s): 248

Problem Description

Coco has a tree, whose vertices are conveniently labeled by 1,2,…,n.

There are m chain on the tree, Each chain has a certain weight. Coco would like to pick out some chains any two of which do not share common vertices.

Find out the maximum sum of the weight Coco can pick

Input

The input consists of several test cases. The first line of input gives the number of test cases T (T<=10).

For each tests:

First line two positive integers n, m.(1<=n,m<=100000)

The following (n - 1) lines contain 2 integers ai bi denoting an edge between vertices ai and bi (1≤ai,bi≤n),

Next m lines each three numbers u, v and val（1≤u，v≤n，0<val<1000）, represent the two end points and the weight of a tree chain.

Output

For each tests:

A single integer, the maximum number of paths.

Sample Input

1
7 3
1 2
1 3
2 4
2 5
3 6
3 7
2 3 4
4 5 3
6 7 3

Sample Output

6

Hint

 Stack expansion program： #pragma comment(linker, "/STACK:1024000000,1024000000")

Author

FZUACM

Source

2015 Multi-University Training Contest 1

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考虑dp，dp[i]表示以i为根的子树的最优值，则

sum[i]=∑j∈son[i] 
d p[j]

容易想到有两种转移

  • (1) dp[i]=sum[i]
  • (2) dp[i]=value[p]+∑sum[k]?∑dp[k] (
     链p的lca是i,k是链上的节点)

    链上求和很容易想到树链剖分，复杂度O(Nlog2N)

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
#include<vector>
#include<iomanip>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (100000+10)
#define MAXM (200000+10)
#define MAXV (1000+10)
#define pb push_back
#define mp make_pair
#pragma comment(linker, "/STACK:1024000000,1024000000")
typedef int ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
struct Chain{
	int u,v,w;
	Chain(){}
	Chain(int _u,int _v,int _w):u(_u),v(_v),w(_w){}
};
vector<Chain> a[MAXN];
int n,m;
int edge[MAXM],Next[MAXM],Pre[MAXN],siz=1;
void addedge(int u,int v)
{
	edge[++siz]=v;
	Next[siz]=Pre[u];
	Pre[u]=siz;
}
void addedge2(int u,int v){addedge(u,v);addedge(v,u);}

bool vis[MAXN];
int cnt,id[MAXN];
int son[MAXN],dep[MAXN],sz[MAXN],top[MAXN],pre[MAXN],q[MAXN];
void build()
{
	MEM(vis) cnt=0; MEM(id)
	MEM(son) MEM(dep) MEM(sz) MEM(top) MEM(pre) MEM(q)
	int r=1;
	vis[dep[1]=q[1]=1]=1;
	For(i,r)
	{
		int u=q[i];
		Forp(u)
		{
			int v=edge[p];
			if (vis[v]) continue; else vis[v]=1;
			dep[ q[++r]=v ]=dep[u]+1;
			pre[v]=u;
		}
	}
	ForD(i,r) {
		sz[pre[q[i]]] += ++sz[q[i]];
		if (sz[son[pre[q[i]]]]<sz[q[i]] ) son[pre[q[i]]] = q[i];
	}
	For(i,r) {
		if (!top[q[i]])
			for(int x=q[i];x;x=son[x]) {
				top[x]=q[i];
				id[x]=++cnt;
			}
	}

}
int lca(int a,int b)
{
	while(1) {
		if (top[a]==top[b]) return dep[a]<=dep[b] ? a:b;
		if (dep[top[a]]<dep[top[b]]) swap(a,b);
		a=pre[top[a]];
	}
}

#define MEM2(a,i) memset(a,i,sizeof(a));
class SegmentTree
{
    ll a[MAXN*4],minv[MAXN*4],sumv[MAXN*4],maxv[MAXN*4],addv[MAXN*4],setv[MAXN*4];
    int n;
public:
    SegmentTree(){MEM(a) MEM(minv) MEM(sumv) MEM(maxv) MEM(addv) MEM2(setv,-1) }
    SegmentTree(int _n):n(_n){MEM(a) MEM(minv) MEM(sumv) MEM(maxv) MEM(addv) MEM2(setv,-1) }
    void mem(int _n)
    {
        n=_n;
        MEM(a) MEM(minv)  MEM(sumv) MEM(maxv) MEM(addv) MEM2(setv,-1)
    }  

    void maintain(int o,int L,int R)
    {

		sumv[o]=maxv[o]=minv[o]=0;
    	if (L<R) //只考虑左右子树
		{
			sumv[o]=sumv[Lson]+sumv[Rson];
			minv[o]=min(minv[Lson],minv[Rson]);
			maxv[o]=max(maxv[Lson],maxv[Rson]);
		} //只考虑add操作
		if (setv[o]>=0) sumv[o]=setv[o]*(R-L+1),minv[o]=maxv[o]=setv[o];

		minv[o]+=addv[o];maxv[o]+=addv[o];sumv[o]+=addv[o]*(R-L+1);
    }

	int y1,y2,v;
	void update(int o,int L,int R) //y1,y2,v
	{
		if (y1<=L&&R<=y2) {
			addv[o]+=v;
		}
		else{
			pushdown(o);
			int M=(R+L)>>1;
			if (y1<=M) update(Lson,L,M); else maintain(Lson,L,M);
			if (M< y2) update(Rson,M+1,R); else maintain(Rson,M+1,R);
		}

		maintain(o,L,R); 

	}
	void update2(int o,int L,int R)
	{
		if (y1<=L&&R<=y2) {
			setv[o]=v;addv[o]=0;
		}
		else{
			pushdown(o);
			int M=(R+L)>>1;
			if (y1<=M) update2(Lson,L,M); else maintain(Lson,L,M); //维护pushodown，再次maintain
			if (M< y2) update2(Rson,M+1,R); else maintain(Rson,M+1,R);
		}

		maintain(o,L,R);
	}

	void pushdown(int o)
	{
		if (setv[o]>=0)
		{
			setv[Lson]=setv[Rson]=setv[o];
			addv[Lson]=addv[Rson]=0;
			setv[o]=-1;
		}
		if (addv[o])
		{
			addv[Lson]+=addv[o];
			addv[Rson]+=addv[o];
			addv[o]=0;
		}
	}
	ll _min,_max,_sum; 

	void query2(int o,int L,int R,ll add)
	{
		if (setv[o]>=0)
		{
			_sum+=(setv[o]+addv[o]+add)*(min(R,y2)-max(L,y1)+1);
			_min=min(_min,setv[o]+addv[o]+add);
			_max=max(_max,setv[o]+addv[o]+add);
		} else if (y1<=L&&R<=y2)
		{
			_sum+=sumv[o]+add*(R-L+1);
			_min=min(_min,minv[o]+add);
			_max=max(_max,maxv[o]+add);
		} else {
		//	pushdown(o);
			int M=(L+R)>>1;
			if (y1<=M) query2(Lson,L,M,add+addv[o]);// else maintain(Lson,L,M);
			if (M< y2) query2(Rson,M+1,R,add+addv[o]);// else maintain(Rson,M+1,R);
		}
		//maintain(o,L,R);
	}

	void query(int o,int L,int R,ll add) //y1,y2
	{
		if (y1<=L&&R<=y2)
		{
			_sum+=sumv[o]+add*(R-L+1);
			_min=min(_min,minv[o]+add);
			_max=max(_max,maxv[o]+add);
		}
		else{
			int M=(R+L)>>1;
			if (y1<=M) query(Lson,L,M,add+addv[o]);
			if (M< y2) query(Rson,M+1,R,add+addv[o]);
		}
	}

	void add(int l,int r,ll v)
	{
		if (l>r) swap(l,r);
		y1=l,y2=r;this->v=v;
		update(1,1,n);
	}
	void set(int l,int r,ll v)
	{
		y1=l,y2=r;this->v=v;
		update2(1,1,n);
	}
	ll ask(int l,int r,int b=1)
	{
		if (l>r) swap(l,r);
		_sum=0,_min=INF,_max=-1;
		y1=l,y2=r;
		query2(1,1,n,0);
		switch(b)
		{
			case 1:return _sum;
			case 2:return _min;
			case 3:return _max;
			default:break;
		}
	}
	void print()
	{
		For(i,n)
			cout<<ask(i,i,1)<<' ';
		cout<<endl;

	}

    //先set后add
}S[2];   //sum & dp

int d[MAXN],s[MAXN];

ll Ask(int a,int b,int f)
{
	ll ans=0;
	while (top[a]^top[b]) {
		if (dep[top[a]]<dep[top[b]]) swap(a,b);
		ans+=S[f].ask(id[top[a]],id[a],1);
		a=pre[top[a]];
	}
	if (dep[a]>dep[b]) swap(a,b);
	ans+=S[f].ask(id[a],id[b],1);
	return ans;
}

void dfs(int u,int fa)
{
	Forp(u)
	{
		int v=edge[p];
		if (v==fa) continue;
		dfs(v,u);
		s[u]+=d[v];
	}
	d[u]=s[u];
	S[0].add(id[u],id[u],s[u]);
	int tot=a[u].size();
	Rep(j,tot)
	{
		Chain t = a[u][j];
		int fee=t.w;
		//
		d[u]=max(d[u],(int)(fee+Ask(t.u,t.v,0)-Ask(t.u,t.v,1)));
	}
	S[1].add(id[u],id[u],d[u]);  

}

int main()
{
//	freopen("hdu5293.in","r",stdin);

	int T;cin>>T;
	while(T--) {
		MEM(edge) MEM(Next) MEM(Pre) siz=1;
		MEM(d) MEM(s)
		For(i,n) a[i].clear();

		cin>>n>>m;
		S[0].mem(n);S[1].mem(n);
		For(i,n-1)
		{
			int u,v;
			scanf("%d%d",&u,&v);
			addedge2(u,v);
		}
		build();

		For(i,m) {
			int u,v ,w;
			scanf("%d%d%d",&u,&v,&w);
			a[lca(u,v)].pb(Chain(u,v,w));
		}
		dfs(1,0);
		printf("%d\n",d[1]);
	}	

	return 0;
}

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时间： 2024-10-23 01:25:50

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