poj3764 The XOR Longest Path【dfs】【Trie树】

The xor-longest Path

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 10038		Accepted: 2040

Description

In an edge-weighted tree, the xor-length of a path p is defined as the xor sum of the weights of edges on p:

⊕ is the xor operator.

We say a path the xor-longest path if it has the largest xor-length. Given an edge-weighted tree with n nodes, can you find the xor-longest path? 　

Input

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 uand v of length w.

Output

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

Sample Input

4
0 1 3
1 2 4
1 3 6

Sample Output

7

Hint

The xor-longest path is 0->1->2, which has length 7 (=3 ⊕ 4)

Source

题意：

给一棵带边权的树，想找得到两个点，他们的路径上的权值异或最小。

思路：

首先我们任意找一个作为根，可以用dfs求出其他节点到根的路径的异或，记为xordis

那么对于树上的任意两个节点i, j，i到j的路径的异或和应该是xordis[i] ^ xordis[j]

因为i到j的路径，相当于i到根，根到j，其中重叠的部分，他们的异或值正好是0

因此这道题就变成了找两点异或值最小，https://www.cnblogs.com/wyboooo/p/9824293.html 和这道题就差不多了

最后还需要注意，search找到的最大值是除根以外的，还需要和xordis比较一下，取较大值。

#include <iostream>
#include <set>
#include <cmath>
#include <stdio.h>
#include <cstring>
#include <algorithm>
#include <map>
using namespace std;
typedef long long LL;
#define inf 0x7f7f7f7f

int n;
const int maxn = 1e5 + 5;
struct edge{
    int v, w;
    int nxt;
}e[maxn * 2];
int head[maxn], tot = 0;
int xordis[maxn];
int trie[maxn * 32 + 5][3], treetot = 1;

void addedge(int u, int v, int w)
{
    e[tot].v = v;
    e[tot].w = w;
    e[tot].nxt = head[u];
    head[u] = tot++;
    e[tot].v = u;
    e[tot].w = w;
    e[tot].nxt = head[v];
    head[v] = tot++;
}

void dfs(int rt, int fa)
{
    for(int i = head[rt]; i != -1; i = e[i].nxt){
        int v = e[i].v;
        if(v == fa)continue;
        xordis[v] = xordis[rt] ^ e[i].w;
        dfs(v, rt);
    }
}

void init()
{
    memset(head, -1, sizeof(head));
    tot = 0;
    memset(xordis, 0, sizeof(xordis));
    memset(trie, 0, sizeof(trie));
}

void insertt(int x)
{
    int p = 1;
    for(int i = 30; i >= 0; i--){
        int ch = x >> i & 1;
        if(trie[p][ch] == 0){
            trie[p][ch] = ++tot;
        }
        p = trie[p][ch];
    }
}

int searchh(int x)
{
    int p = 1, ans = 0;
    for(int i = 30; i >= 0; i--){
        int ch = x >> i & 1;
        if(trie[p][ch ^ 1]){
            p = trie[p][ch ^ 1];
            ans |= 1 << i;
        }
        else{
            p = trie[p][ch];
        }
    }
    return ans;
}

int main()
{
    while(scanf("%d", &n) != EOF){
        init();
        for(int i = 0; i < n - 1; i++){
            int u, v, w;
            scanf("%d%d%d", &u, &v, &w);
            addedge(u, v, w);
        }
        dfs(0, -1);

        /*for(int i = 0; i < n; i++){
            printf("%d
", xordis[i]);
        }*/

        int ans = 0;
        for(int i = 1; i < n; i++){
            insertt(xordis[i]);
            //cout<<searchh(xordis[i])<<endl;
            ans = max(ans, searchh(xordis[i]));
            ans = max(ans, xordis[i]);
        }
        printf("%d
", ans);
    }
    return 0;
}