hdu 5029树链剖分+线段树

Relief grain

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 100000/100000 K (Java/Others)
Total Submission(s): 437    Accepted Submission(s): 99

Problem Description

The soil is cracking up because of the drought and the rabbit kingdom is facing a serious famine. The RRC(Rabbit Red Cross) organizes the distribution of relief grain in the disaster area.

We can regard the kingdom as a tree with n nodes and each node stands for a village. The distribution of the relief grain is divided into m phases. For each phases, the RRC will choose a path of the tree and distribute some relief grain of a certain type for every village located in the path.

There are many types of grains. The RRC wants to figure out which type of grain is distributed the most times in every village.

 

Input

The input consists of at most 25 test cases.

For each test case, the first line contains two integer n and m indicating the number of villages and the number of phases.

The following n-1 lines describe the tree. Each of the lines contains two integer x and y indicating that there is an edge between the x-th village and the y-th village.
  
The following m lines describe the phases. Each line contains three integer x, y and z indicating that there is a distribution in the path from x-th village to y-th village with grain of type z. (1 <= n <= 100000, 0 <= m <= 100000, 1 <= x <= n, 1 <= y <= n, 1 <= z <= 100000)

The input ends by n = 0 and m = 0.

 

Output

For each test case, output n integers. The i-th integer denotes the type that is distributed the most times in the i-th village. If there are multiple types which have the same times of distribution, output the minimal one. If there is no relief grain in a village, just output 0.

 

Sample Input

  

   2 4
1 2
1 1 1
1 2 2
2 2 2
2 2 1
5 3
1 2
3 1
3 4
5 3
2 3 3
1 5 2
3 3 3
0 0
  

 

Sample Output

  

   1
2
2
3
3
0
2


   

    

     Hint
    
  For the first test case, the relief grain in the 1st village is {1, 2}, and the relief grain in the 2nd village is {1, 2, 2}.
   
    
  

 

Source

2014 ACM/ICPC Asia Regional Guangzhou Online

#pragma comment(linker, "/STACK:102400000,102400000")
#include<stdio.h>
#include<algorithm>
#include<string.h>
#define N 100010
#include<vector>
using namespace std;
vector<int>add[N],del[N];
struct node
{
   int u,v,next;
}bian[N*2];
struct pp
{
   int x,y;
   int col;//区间当中数目最多的那个数
   int num;//区间当中最大的重复次数
}a[N*4];
int e,head[N],sz[N],dep[N],son[N],father[N],ti[N],top[N],id,dui[N],ans[N];
void adge(int u,int v)
{
    bian[e].u=u;
    bian[e].v=v;
    bian[e].next=head[u];
    head[u]=e++;
}
int max(int a,int b)
{
 return a>b?a:b;
}
void dfs1(int u,int fa)
{
  int i,v;
   sz[u]=1; dep[u]=dep[fa]+1; son[u]=0; father[u]=fa;
   for(i=head[u];i!=-1;i=bian[i].next)
   {
     v=bian[i].v;
     if(v==fa)  continue;
     dfs1(v,u);
     sz[u]+=sz[v];
     if(sz[son[u]]<sz[v])
        son[u]=v;
   }
}
void dfs2(int u,int fa)
{
    int i,v;
    ti[u]=id++;
    top[u]=fa;
    if(son[u]!=0)
         dfs2(son[u],fa);
    for(i=head[u];i!=-1;i=bian[i].next)
    {
      v=bian[i].v;
      if(v==father[u]||v==son[u])
           continue;
        dfs2(v,v);
    }
}
void cal(int u,int v,int z)
{
    while(top[u]!=top[v])
    {
        if(dep[top[u]]>dep[top[v]])
           swap(u,v);
        add[ti[top[v]]].push_back(z);
        del[ti[v]].push_back(z);
        v=father[top[v]];
    }
    if(ti[u]>ti[v])
      swap(u,v);
    add[ti[u]].push_back(z);
    del[ti[v]].push_back(z);
}
void pushdown(int t)
{
   int temp=t<<1;
   if(a[temp].num>=a[temp+1].num)
   {
     a[t].col=a[temp].col;
     a[t].num=a[temp].num;
   }
   else
   {
      a[t].col=a[temp+1].col;
      a[t].num=a[temp+1].num;
   }
}
void update(int k,int t,int p)
{
    if(a[t].x==a[t].y)
    {
       a[t].num+=p;
       return ;
    }
    int mid=(a[t].x+a[t].y)>>1,temp=t<<1;
    if(k<=mid)
       update(k,temp,p);
    else
       update(k,temp+1,p);
    pushdown(t);
}
void build(int x,int y,int t)
{
     a[t].x=x;
     a[t].y=y;
     if(x==y)
     {
        a[t].col=x;
        a[t].num=0;
        return ;
     }
     int mid=(x+y)>>1,temp=t<<1;
     build(x,mid,temp);
     build(mid+1,y,temp+1);
     pushdown(t);
}
int main()
{
  int n,m,i,u,v,z,cnt,j,sx;
  //freopen("a.txt","r",stdin);
  while(scanf("%d%d",&n,&m)!=EOF)
  {
     if(n==0&&m==0)  break;
     memset(head,-1,sizeof(head));
     e=0;
     for(i=1;i<n;i++)
     {
        scanf("%d%d",&u,&v);
        adge(u,v);
        adge(v,u);
     }
     dep[1]=0; id=1; sz[0]=0;
     dfs1(1,1);
     dfs2(1,1);
     /*for(i=1;i<=n;i++)
     {
         printf("i=%d ",i);
         printf("dep=%d ",dep[i]);
         printf("sz=%d ",sz[i]);
         printf("son=%d ",son[i]);
         printf("father=%d ",father[i]);
         printf("ti=%d ",ti[i]);
         printf("top=%d ",top[i]);
         printf("\n");
     }*/
     for(i=1;i<=n;i++)
     {
        dui[ti[i]]=i;
     }
     cnt=0;
     for(i=1;i<=m;i++)
     {
        scanf("%d%d%d",&u,&v,&z);
        cal(u,v,z);
        cnt=max(cnt,z);
     }
     build(0,cnt,1);
     for(i=1;i<=n;i++)
     {
         sx=add[i].size();
         for(j=0;j<sx;j++)
         {
            update(add[i][j],1,1);
         }
         ans[dui[i]]=a[1].col;
         sx=del[i].size();
         for(j=0;j<sx;j++)
         {
            update(del[i][j],1,-1);
         }
         add[i].clear();
         del[i].clear();
     }
     for(i=1;i<=n;i++)
       printf("%d\n",ans[i]);
  }
  return 0;
}