【bzoj1954】Pku3764 The xor-longest Path Trie树

题目描述

 给定一棵n个点的带权树,求树上最长的异或和路径

输入

The input contains several test cases. The first line of each test case contains an integer n(1<=n<=100000), The following n-1 lines each contains three integers u(0 <= u < n),v(0 <= v < n),w(0 <= w < 2^31), which means there is an edge between node u and v of length w.

输出

For each test case output the xor-length of the xor-longest path.

样例输入

4
1 2 3
2 3 4
2 4 6

样例输出

7


题解

Trie树

由于x^x=0,所以树上x和y之间路径的异或和 = x到根路径的异或和 xor y到根路径的异或和。

所以我们先对整棵树进行dfs,求出每个节点到根的路径异或和dis,并加入到Trie树。

然后枚举树上的节点,在Trie树中贪心查询与它异或和最大的数,并加到答案中即可。

#include <cstdio>
#include <algorithm>
#define N 100010
using namespace std;
int head[N] , to[N << 1] , len[N << 1] , next[N << 1] , cnt , v[N] , c[N * 30][2] , tot;
void add(int x , int y , int z)
{
	to[++cnt] = y , len[cnt] = z , next[cnt] = head[x] , head[x] = cnt;
}
void dfs(int x , int fa)
{
	int i;
	for(i = head[x] ; i ; i = next[i])
		if(to[i] != fa)
			v[to[i]] = v[x] ^ len[i] , dfs(to[i] , x);
}
void insert(int x)
{
	int i , p = 0;
	bool t;
	for(i = 1 << 30 ; i ; i >>= 1)
	{
		t = x & i;
		if(!c[p][t]) c[p][t] = ++tot;
		p = c[p][t];
	}
}
int query(int x)
{
	int i , p = 0 , ans = 0;
	bool t;
	for(i = 1 << 30 ; i ; i >>= 1)
	{
		t = x & i;
		if(c[p][t ^ 1]) ans += i , p = c[p][t ^ 1];
		else p = c[p][t];
	}
	return ans;
}
int main()
{
	int n , i , x , y , z , ans = 0;
	scanf("%d" , &n);
	for(i = 1 ; i < n ; i ++ ) scanf("%d%d%d" , &x , &y , &z) , add(x , y , z) , add(y , x , z);
	dfs(1 , 0);
	for(i = 1 ; i <= n ; i ++ ) insert(v[i]);
	for(i = 1 ; i <= n ; i ++ ) ans = max(ans , query(v[i]));
	printf("%d\n" , ans);
	return 0;
}

 

 

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posted @ 2017-06-24 10:13  GXZlegend  阅读(324)  评论(0编辑  收藏  举报