hdu 5416 CRB and Tree 2015多校联合训练赛#10 枚举

CRB and Tree

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 628    Accepted Submission(s): 198

Problem Description

CRB has a tree, whose vertices are labeled by 1, 2, …,  N . They are connected by  N  – 1 edges. Each edge has a weight.
For any two vertices  u  and  v (possibly equal),  f(u,v)  is xor(exclusive-or) sum of weights of all edges on the path from  u  to  v .
CRB’s task is for given  s , to calculate the number of unordered pairs  (u,v)  such that  f(u,v) = s . Can you help him?

 

Input

There are multiple test cases. The first line of input contains an integer  T , indicating the number of test cases. For each test case:
The first line contains an integer  N  denoting the number of vertices.
Each of the next  N  - 1 lines contains three space separated integers  a ,  b  and  c  denoting an edge between  a  and  b , whose weight is  c .
The next line contains an integer  Q  denoting the number of queries.
Each of the next  Q  lines contains a single integer  s .
1 ≤  T  ≤ 25
1 ≤  N  ≤  105
1 ≤  Q  ≤ 10
1 ≤  a ,  b  ≤  N
0 ≤  c ,  s  ≤  105
It is guaranteed that given edges form a tree.

 

Output

For each query, output one line containing the answer.

 

Sample Input

   
   
   
   

    
    
    
    1
3
1 2 1
2 3 2
3
2
3
4
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    1
1
0

    
    
    
    
 
     
 
      Hint 
     
 For the first query, (2, 3) is the only pair that f(u, v) = 2. For the second query, (1, 3) is the only one. For the third query, there are no pair (u, v) such that f(u, v) = 4. 
    
    
    
    
     
   
   
   
   

 

Author

KUT（DPRK）

 

Source

2015 Multi-University Training Contest 10

求无序的u，v，使得u到v的路径的异或和为s。

从任意一点dfs，记录点到跟的路径上的异或和。

对于每个询问枚举异或值，然后相乘累加。

因为算出的是有序的结果，最后要除以2.

并且对于s = 0时，由于u,u只算了一次，要加起来。

#pragma comment(linker,"/STACK:102400000,102400000")
#include<iostream>
#include<vector>
#include<cstring>
#include<algorithm>
#include<cstdio>
using namespace std;
struct Node{
    int v,d;
    Node (int _v=0,int _d=0):v(_v),d(_d){}
};
#define maxn 100007
vector<Node> head[maxn];
int ans[2*maxn];
void dfs(int u,int f,int n){
    ans[n]++;
    for(int i = 0 ;i < head[u].size() ;i++){
        if(head[u][i].v != f){
            dfs(head[u][i].v,u,n^head[u][i].d);
        }
    }
}
int main(){
    int t,n,q,a,b,c,s;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        for(int i = 0;i <= n; i++)
            head[i].clear();
        for(int i = 1;i < n; i++){
            scanf("%d%d%d",&a,&b,&s);
            head[a].push_back(Node(b,s));
            head[b].push_back(Node(a,s));
        }
        scanf("%d",&q);
        memset(ans,0,sizeof(ans));
        dfs(1,0,0);
        while(q--){
            scanf("%d",&s);
            long long res = 0;
            for(int i = 0;i < (1<<17); i++){
                if(ans[i] == 0) continue;
                res += 1LL*ans[i]*ans[i^s];
            }
            if(s == 0) res += n;
            printf("%I64d\n",res/2);
        }
    }
    return 0;
}