HDU 5416 CRB and Tree （2015多校第10场）

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CRB and Tree

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 481    Accepted Submission(s): 151

Problem Description

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

 

Input

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

 

Output

For each query, output one line containing the answer.

 

Sample Input

131 2 12 3 23234

 

Sample Output

110
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）

解题思路：

因为异或是可逆的，因此从前到后记录前缀异或和，用hash表记录每个值出现的次数，每次只需要加上x ^ sum[v]出现的次数即可。因为此时，u到v的异或和就是x。

#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>#include <vector>#include <queue>#include <set>#include <map>#include <stack>#include <algorithm>#define LL long long using namespace std;const int MAXN = 100000 + 10;struct Edge{int to, next, w;}edge[MAXN<<1];int tot, head[MAXN];int read(){int res = 0, f = 1; char ch = getchar();while(ch < '0' || ch > '9'){if(ch == '-') f *= -1; ch = getchar();}while(ch >= '0' && ch <= '9'){res = res * 10 + ch - '0'; ch = getchar();}return res;}void init(){tot = 0;memset(head, -1, sizeof(head));}void addedge(int u, int v, int w){edge[tot].to = v;edge[tot].w = w;edge[tot].next = head[u];head[u] = tot++;}int N, Q;int vis[MAXN], st[MAXN<<2], op;LL ans;void dfs(int u, int x){vis[u] = 1; st[x]++;ans += st[op ^ x];for(int i=head[u];i!=-1;i=edge[i].next){int v = edge[i].to, w = edge[i].w;if(!vis[v]){dfs(v, x ^ w);}}}int main(){int T;scanf("%d", &T);while(T--){N = read();int u, v, w;init();for(int i=1;i<N;i++){u = read(); v = read(); w = read();addedge(u, v, w);addedge(v, u, w);}scanf("%d", &Q);while(Q--){op = read();memset(vis, 0, sizeof(vis));memset(st, 0, sizeof(st));ans = 0;dfs(1, 0);printf("%I64d\n", ans);}}return 0;}