HDU6074 Phone Call（并查集，lca）

Phone Call

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 395    Accepted Submission(s): 163

Problem Description

There are  n  houses in Bytetown, labeled by  1,2,...,n . In each house, there is a person living here. Little Q lives in house  1 . There are  n−1  bidirectional streets connecting these houses, forming a tree structure. In this problem,  S(u,v)  denotes the house set containing all the houses on the shortest path from house  u  to house  v .

The Bytetown's phone line network consists of  m  different lines. The  i -th line can be expressed as  5  integers  ai,bi,ci,di,wi , which means for every two different houses  u  and  v  from set  S(ai,bi)∪S(ci,di) ,  u  and  v  can have a phone call costing  wi  dollars.



Picture from Wikimedia Commons


Little Q is now planning to hold a big party in his house, so he wants to make as many as possible people known. Everyone known the message can make several phone calls to others to spread the message, but nobody can leave his house.

Please write a program to figure out the maximum number of people that can join the party and the minimum total cost to reach that maximum number. Little Q should be counted in the answer.

 

Input

The first line of the input contains an integer  T(1≤T≤15) , denoting the number of test cases.

In each test case, there are  2  integers  n,m(1≤n,m≤100000)  in the first line, denoting the number of houses and phone lines.

For the next  n−1  lines, each line contains two integers  u  and  v , denoting a bidirectional edge between node  u  and  v .

For the next  m  lines, each line contains  5  integers  ai,bi,ci,di,wi(1≤ai,bi,ci,di≤n,1≤wi≤109) , denoting a phone line.

 

Output

For each test case, print a single line containing two integers, denoting the maximum number of people that can join the party and the minimum total cost to reach that maximum number.

 

Sample Input

  

   1
5 2
1 2
1 3
2 4
2 5
1 3 2 4 100
2 2 4 2 10
  

 

Sample Output

  

   4 210

   

    

     Hint
    
Step 1 : 1 make a phone call to 2 using line 1, the cost is 100.
Step 2 : 1 make a phone call to 3 using line 1, the cost is 100.
Step 3 : 2 make a phone call to 4 using line 2, the cost is 10.

   
    
  

 

Source

2017 Multi-University Training Contest - Team 4 

 

Recommend

liuyiding

 

题意：有一颗树，每次给四个点，a,b,c,d.告诉你a到b路径上的节点可以相互到达同时和同时c到d路径上的节点可以相互到达。且花费为w。

求出1最多可以到达几个节点，并且求出最小花费。

整个过程类似最小生成树的合并过程，所以按照花费从小到大排序，首先把两个端点和其LCA合并，在把两个LCA合并。

并查集维护这个过程。

#include <map>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
using namespace std;
const int MAXN=400000+10;
const int mod=1e9+7;
struct edge{
    int to,next;
}e[MAXN<<1];
int head[MAXN];
int tot=0;
void add(int u,int v){
    e[tot].to=v;
    e[tot].next=head[u];
    head[u]=tot++;
}

int dep[MAXN];
int fir[MAXN];
int que[MAXN<<1];
int p[MAXN];
int st[MAXN][20];
int vis[MAXN];
int cnt=0;

void dfs(int u,int d){
    vis[u]=1;
    que[++cnt]=u;fir[u]=cnt;dep[cnt]=d;
    for(int k=head[u];k!=-1;k=e[k].next){
        int v=e[k].to;
        if(!vis[v]){
            p[v]=u;
            dfs(v,d+1);
            que[++cnt]=u;dep[cnt]=d;
        }
    }
}
void ST(int n){
    for(int i=1;i<=n;i++)
        st[i][0]=i;
    for(int j=1;(1<<j)<=n;j++){
        for(int i=1;i+(1<<j)-1<=n;i++){
            int a=st[i][j-1],b=st[i+(1<<(j-1))][j-1];
            st[i][j]=dep[a]<dep[b]?a:b;
        }
    }
}
int RMQ(int l,int r){
    int k=0;
    while((1<<(k+1))<=r-l+1){
        k++;
    }
    int a=st[l][k],b=st[r-(1<<k)+1][k];
    return dep[a]<dep[b]?a:b;
}

int lca(int u,int v){
    int x=fir[u],y=fir[v];
    if(x>y)
        swap(x,y);
    int res=RMQ(x,y);
    return que[res];
}


struct node{
    int a,b,c,d;
    long long w;
}s[MAXN];
int cmp(node x,node y){
    return x.w<y.w;
}

int par[MAXN],up[MAXN],cn[MAXN];
long long w[MAXN];
int findFa(int x){
    if(par[x]==x)
        return x;
    else
        return par[x]=findFa(par[x]);
}
int findUp(int x){
    if(up[x]==x)
        return x;
    else
        return up[x]=findUp(up[x]);
}

void Union(int x,int y,long long cost){
    x=findFa(x),y=findFa(y);
    if(x==y)
        return;
    par[x]=y;
    cn[y]+=cn[x];
    w[y]+=w[x]+cost;
}

void merge(int u,int v,long long cost){
    while(1){
        u=findUp(u);
        if(dep[fir[u]]<=dep[fir[v]]){
            //cout<<u<<" "<<v<<" "<<dep[fir[u]]<<" "<<dep[que[fir[v]]]<<endl;
            return;
        }
        Union(u,p[u],cost);
        up[u]=p[u];
    }
}

void solve(node x){
    int LCA=lca(x.a,x.b);
    merge(x.a,LCA,x.w);
    merge(x.b,LCA,x.w);
    LCA=lca(x.c,x.d);
    merge(x.c,LCA,x.w);
    merge(x.d,LCA,x.w);
    Union(x.a,x.c,x.w);
}

void init(int n){
    tot=0;
    cnt=0;
    p[1]=0;
    memset(head,-1,sizeof head);
    memset(vis,0,sizeof vis);
    for(int i=0;i<=n;i++){
        par[i]=up[i]=i;
        cn[i]=1;
        w[i]=0;
    }
}

int main(){
    int t;
    //freopen("1008.in","r",stdin);
    cin>>t;
    while(t--){
        int n,m;
        scanf("%d%d",&n,&m);
        init(n);
        for(int i=0;i<n-1;i++){
            int u,v;
            scanf("%d%d",&u,&v);
            add(u,v);
            add(v,u);
        }
        dfs(1,0);
        ST(2*n-1);
        for(int i=0;i<m;i++){
            scanf("%d%d%d%d%lld",&s[i].a,&s[i].b,&s[i].c,&s[i].d,&s[i].w);
        }
        sort(s,s+m,cmp);
        for(int i=0;i<m;i++){
            solve(s[i]);
        }
        printf("%d %lld\n",cn[findFa(1)],w[findFa(1)]);
    }
}