J

Relief grain

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 100000/100000 K (Java/Others)
Total Submission(s): 3246    Accepted Submission(s): 955

题目链接

http://acm.hdu.edu.cn/showproblem.php?pid=5029

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

 

题意

给你一棵树，每次可以给一条路径分配一个值，值可以叠加，最后输出每个点被分配最多的那个值。

 

题解

我们先考虑一个问题，如果是在一个序列里，每次给你一个区间(l,r),问每个点最后被多少个区间覆盖。

这道题是一道比较经典的扫描线的题，如果不会我来讲讲：

对于每个区间(l,r),我们把他拆成两个点，(l,1) 和（r+1,-1)，(1代码左端点，-1表示右端点),并将其排序。然后假想有一条扫描线从左扫到右，每次碰到一个左端点答案加一，碰到右端点答案减一。

边扫边统计答案，答案就是到当前点，有多少个左端点还没有碰到右端点。

这样可能有点抽象自己可以模拟几遍，应该还是能想出来的。

 

那么和这道题有什么关系呢？　当然有关系啦，不然我为什么会将呢？

１．我们加一颗权值线段树，对于每个区间我们增加了一维权值来确定，即(l,r,val)表示l~r赋值为val,那么还是如上一样扫描，然后答案加一的步骤改成给线段树点val加上1，

这样每个点的答案就是线段树中最大值的位置。

 

２．但是这是在树上，所以树链剖分将他变成一个连续序列即可。

 

代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define N 200050
struct Edge{int from,to,s;}edges[N<<1];
struct Query
{
  int x,val,id;
  bool operator <(const Query&b)const
  {return x<b.x;}
}a[N<<2];
struct Tree{int l,r,mx,wmx;}tr[N<<2];
int n,m,num,ans[N];
int tot,last[N];
int cnt,fa[N],dp[N],size[N],son[N],rk[N],kth[N],top[N];
template<typename T>void read(T&x)
{
  ll k=0; char c=getchar();
  x=0;
  while(!isdigit(c)&&c!=EOF)k^=c=='-',c=getchar();
  if (c==EOF)exit(0);
  while(isdigit(c))x=x*10+c-'0',c=getchar();
  x=k?-x:x;
}
void read_char(char &c)
{while(!isalpha(c=getchar())&&c!=EOF);}
void AddEdge(int x,int y)
{
  edges[++tot]=Edge{x,y,last[x]};
  last[x]=tot;
}
void dfs1(int x,int pre)
{
  fa[x]=pre;
  dp[x]=dp[pre]+1;
  size[x]=1;
  son[x]=0;
  for(int i=last[x];i;i=edges[i].s)
    {
      Edge &e=edges[i];
      if (e.to==pre)continue;
      dfs1(e.to,x);
      size[x]+=size[e.to];
      if (size[e.to]>size[son[x]])
    son[x]=e.to;
    }
}
void dfs2(int x,int y)
{
  rk[x]=++cnt;
  kth[cnt]=x;
  top[x]=y;
  if (son[x]==0)return;
  dfs2(son[x],y);
  for(int i=last[x];i;i=edges[i].s)
    {
      Edge e=edges[i];
      if (e.to==fa[x]||e.to==son[x])continue;
      dfs2(e.to,e.to);
    }
}
void push_up(Tree &c,Tree a,Tree b)
{
  c.mx=max(a.mx,b.mx);
  c.wmx=c.mx==a.mx?a.wmx:b.wmx;
}
void bt(int x,int l,int r)
{
  tr[x]=Tree{l,r,0,0};
  if (l==r){tr[x].wmx=l;return;}
  int mid=(l+r)>>1;
  bt(x<<1,l,mid);
  bt(x<<1|1,mid+1,r);
}
void update(int x,int p,int tt)
{
  if (p<=tr[x].l&&tr[x].r<=p)
    {
      tr[x].mx+=tt;
      tr[x].wmx=tr[x].l;
      return;
    }
  int mid=(tr[x].l+tr[x].r)>>1;
  if (p<=mid)update(x<<1,p,tt);
  if (mid<p)update(x<<1|1,p,tt);
  push_up(tr[x],tr[x<<1],tr[x<<1|1]);
}
void change(int x,int y,int tt)
{
  int fx=top[x],fy=top[y];
  while(fx!=fy)
    {
      if (dp[fx]<dp[fy])swap(x,y),swap(fx,fy);
      a[++num]=Query{rk[fx],tt,1};
      a[++num]=Query{rk[x]+1,tt,-1};
      x=fa[fx];fx=top[x];
    }
  if (dp[x]<dp[y])swap(x,y);
  a[++num]=Query{rk[y],tt,1};
  a[++num]=Query{rk[x]+1,tt,-1};
}
void work()
{
  read(n); read(m);
  if (n==0)exit(0);
  for(int i=1;i<=n-1;i++)
    {
      int x,y;
      read(x); read(y);
      AddEdge(x,y);
      AddEdge(y,x);
    }
  dfs1(1,0);
  dfs2(1,1);
  bt(1,0,100005);
  for(int i=1;i<=m;i++)
    {
      int x,y,tt;
      read(x); read(y); read(tt);
      change(x,y,tt);
    }
  sort(a+1,a+num+1);
  for(int i=1;i<=num-1;i++)
    {
      int l=a[i].x;
      update(1,a[i].val,a[i].id);
      while(a[i+1].x==a[i].x)
    //update(1,a[++i].val,a[i].id);
    i++,update(1,a[i].val,a[i].id);
      
      for(int j=l;j<a[i+1].x;j++)
    ans[kth[j]]=tr[1].wmx;
    }
  for(int i=1;i<=n;i++)printf("%d
",ans[i]);
}
void clear()
{
  tot=0; cnt=0; num=0;
  memset(last,0,sizeof(last));
  memset(ans,0,sizeof(ans));
}
int main()
{
#ifndef ONLINE_JUDGE
  freopen("aa.in","r",stdin);
#endif
  while(1)
    {
      clear();
      work();
    }
}