本文介绍了一种结合树链剖分与动态规划解决特定树形结构问题的方法。该问题要求从一棵树中选择若干条不相交的链，使得这些链的权值之和最大。文章详细解释了树链剖分的概念及其应用，并通过具体实现展示了如何高效地求解此类问题。
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
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.
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.
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
6HintStack expansion program： #pragma comment(linker, "/STACK:1024000000,1024000000")
Author
FZUACM
Source
Recommend
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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