Tree chain problem

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1752 Accepted Submission(s): 561

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

Hint

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

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn=1e5+88;
int fa[maxn],dep[maxn],size[maxn],pos[maxn],bl[maxn],head[maxn];
int sum[maxn];
int dp[maxn],su[maxn],n,m;
vector<int>G[maxn];
struct node{
   int to,next;
}edge[maxn<<1];
struct cst{
   int x,y,z;
}road[maxn];
int tot,sz;
void init(){
   tot=sz=0;
   memset(sum,0,sizeof(sum));
   memset(head,-1,sizeof(head));
   for(int i=1;i<=n;++i) G[i].clear();
}
void add(int u,int v) {
   edge[tot].to=v;
   edge[tot].next=head[u];
   head[u]=tot++;
}
void sadd(int u,int val) {
   for( ; u<=n;u+=u&(-u))
    sum[u]+=val;
}
int getsum(int u) {
    int ret=0;
   for(;u;u-=u&(-u))
    ret+=sum[u];
   return ret;
}
void dfs1(int x){
    size[x]=1;
    for(int i=head[x];i+1;i=edge[i].next){
        int v=edge[i].to;
        if(v==fa[x]) continue;
        fa[v]=x;
        dep[v]=dep[x]+1;
        dfs1(v);
        size[v]+=size[x];
    }
}
void dfs2(int x,int chain)
{
    bl[x]=chain;
    pos[x]=++sz;
    int k=0;
    for(int i=head[x];i+1;i=edge[i].next){
        int v=edge[i].to;
        if(dep[v]>dep[x]&&size[v]>size[k])
            k=v;
    }
    if(!k) return;
    dfs2(k,chain);
    for(int i=head[x];i+1;i=edge[i].next)
        if(dep[edge[i].to]>dep[x]&&edge[i].to!=k)
        dfs2(edge[i].to,edge[i].to);
}
int LCA(int x,int y){
   while(bl[x]!=bl[y]) {
    if(dep[bl[x]]<dep[bl[y]]) swap(x,y);
    x=fa[bl[x]];
   }
   if(pos[x]>pos[y]) swap(x,y);
   return x;
}
int query(int x,int y){
    int ret=0;
   while(bl[x]!=bl[y]){
    if(dep[bl[x]]<dep[bl[y]]) swap(x,y);
    ret+=getsum(pos[x])-getsum(pos[bl[x]]-1);
    x=fa[bl[x]];
   }
   if(dep[x]>dep[y]) swap(x,y);
   ret+=getsum(pos[y])-getsum(pos[x]-1);
   return ret;
}
void solve(int u){
   su[u]=0;
   for(int i=head[u];i+1;i=edge[i].next){
    int v=edge[i].to;
    if(v==fa[u]) continue;
    solve(v);
    su[u]+=dp[v];
   }
   dp[u]=su[u];
   for(int i=0;i<(int)G[u].size();++i){
    int v=G[u][i];
    dp[u]=max(dp[u],query(road[v].x,road[v].y)+su[u]+road[v].z);
   }
   sadd(pos[u],su[u]-dp[u]);
}
int main(){
   int t,u,v;
   for(scanf("%d",&t);t--;){
    scanf("%d%d",&n,&m);
    init();
    for(int i=1;i<n;++i){
        scanf("%d%d",&u,&v);
        add(u,v);
        add(v,u);
    }
    dfs1(1);
    dfs2(1,1);
    for(int i=1;i<=m;++i)
    {
        scanf("%d%d%d",&road[i].x,&road[i].y,&road[i].z);
        G[LCA(road[i].x,road[i].y)].push_back(i);
    }
    solve(1);
    printf("%d ",dp[1]);
   }
}