HDU5156（离线tarjan+打标记）

最新推荐文章于 2017-07-19 20:08:00 发布

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最新推荐文章于 2017-07-19 20:08:00 发布

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分类专栏： LCA 文章标签： ACM hdu bestcoder

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Harry and Christmas tree

Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 213 Accepted Submission(s): 53

Problem Description

In Christmas Day, Harry got a Christmas tree. The tree has n nodes. And you can assume the tree rooted at node 1. Harry found that there are some gifts with different colors on some nodes. Then Harry began to count the gifts on the tree. Every time Harrty counted a gift, he written down a pair (a, b) on the paper. Where a indecates which node this gift located, and b indecates the color of this gift. After Harry finished all his counting, for each node x, he wanted to know the number of different kinds of colors that all gifts located on the subtree rooted at node x had.

Input

There are several test cases,you should process to the end of file.
For each test case, there are two integers n m on the first line, n indecates the number of the nodes, m indecates the number of countings.

1≤n≤50000,1≤m≤500000
The next following n-1 lines. Each line contains two integers a b, it means there is an edge between a and b.

1≤a,b≤n,a≠b .
The next following m lines. Each line contains two integers a b, indecates the countings.

1≤a≤n,1≤b≤100000 .

Output

For each test case output one line contains n numbers, where a[i] indecates the number of different kinds of colors that all gifts located on the subtree rooted at node i had.

Sample Input

Sample Output

   
    2 0 0 1 0

    
     
      Hint
     
Make your program running as fast as possible.

Source

BestCoder Round #25

Recommend

heyang | We have carefully selected several similar problems for you: 5157 5154 5153 5152 5151

题意：RT

思路：大概就是打标记，每个点u对从u到树根1上的路径上点贡献为1，在u上+1就好了，这样如果u和v的颜色一样，

那么就会对u和v的最近公共祖先f到树根1上的路径上的点算重复一次，所以只需在f上-1就好了

唯一的坑点是，求最近公共祖先要用离线Tarjan来求，我的这种写法还要加输入挂，否则照样T = =

#pragma comment(linker,"/STACK:102400000,102400000")
#include 
  
   
#include 
   
    
#include 
    
     
#include 
     
      
#include 
      
       
#include 
       
         using namespace std; const int MAXN = 50005; vector
        
          b[MAXN<<1]; vector
         
           a[MAXN]; int head[MAXN]; int edge[MAXN<<1]; int next[MAXN<<1]; int esz; int ans[MAXN]; int n,m; int head1[MAXN]; int edge1[500005]; int next1[500005]; int esz1; int Fa[MAXN]; int vis[MAXN]; void init() { for(int i=1;i<=n;i++){ vis[i]=0; Fa[i]=i; head1[i]=head[i]=-1; ans[i]=0; } esz=esz1=0; } void addedge(int u,int v) { edge[esz]=v; next[esz]=head[u]; head[u]=esz++; } void dfs(int u,int p) { int L=a[u].size(); for(int i=0;i
          
           0)a[i].clear(); for(int i=1;i