CRB and Tree Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 2112    Accepted Submission(s): 635 Problem Description CRB has a tree, whose vertices are labeled by 1, 2, …, N. They are connected…

题目链接: Hdu 5416 CRB and Tree 题目描述: 给一棵树有n个节点,树上的每条边都有一个权值.f(u,v)代表从u到v路径上所有边权的异或值,问满足f(u,v)==m的(u, v)有多少中情况(u, v有可能相同)? 解题思路: 由于xor的特殊性质.x^x=0,对于求f(u, v) == f(u, 1) ^ f(1, u). 又因为x^y == z可以推出x^z == y,对于f(u, 1) ^ f(1, v) == m可以转化为m ^ f(1, v) == f(u, 1)…

题目大意: T, T组测试数据 给你一个n有n个点,下标是从 1 开始的.这个是一棵树,然后下面是n-1条边, 每条边的信息是 s,e,w 代表 s-e的权值是w 然后是一个Q代表Q次询问. 每次询问是一个数字a,  代表 所有路径想异或后的值是 a的路径有多少条. (注:  自己到自己的异或值是0,若a是0要考虑自己到自己)   题目分析: 从根节点开始进行一遍DFS求出所有点到 跟节点的异或值.也就得出来了根节点到达其他任何一个点的异或值. 比如 a 是跟节点到A的异或值, b是根节点到B的…

CRB and Tree Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 967    Accepted Submission(s): 308 Problem Description CRB has a tree, whose vertices are labeled by 1, 2, …, N. They are connected b…

题意:给一棵n个节点的树(无向边),有q个询问,每个询问有一个值s,问有多少点对(u,v)的xor和为s? 注意:(u,v)和(v,u)只算一次.而且u=v也是合法的. 思路:任意点对之间的路径肯定经过LCA的,但是如果如果知道某个点t到根的路径xor和为e,那么就能够得知 x^e=s中的x应该是多少,那就算有多少点到根的xor和为x即可.e是表示t到根的,所以而x也是其他点到根的路径xor和,两个点他们的LCA到根这段会被算2次,那么xor就为0了. (1)DFS算出每个点到根的路径xor和,…

CRB and Tree                                       Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Problem Description CRB has a tree, whose vertices are labeled by 1, 2, …, N. They are connected by N – 1 edges. Ea…

CRB and Tree                                                             Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)                                                                                            To…

欢迎參加--每周六晚的BestCoder(有米!) 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,…

#include<bits/stdc++.h> using namespace std; #define for(i,a,b) for(int i=a;i<=b;++i) //T,N,Q,,u,v,w,s,tree[maxn][?],val[maxn],cashbook[maxn*2],ans const int maxn=1e5+5; struct node { int i,v; node(){} node(int a,int b){i=a,v=b;} }; vector<nod…

Saving HDU Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7194    Accepted Submission(s): 3345 Problem Description 话说上回讲到海东集团面临内外交困,公司的元老也只剩下XHD夫妇二人了.显然,作为多年拼搏的商人,XHD不会坐以待毙的.  一天,当他正在苦思冥想解困良策的…