POJ 3764 The xor-longest Path(xor trie)

题意:

$N\le 10^5个点的树,A_i < 2^{31}$
$路径异或和:=路径上所有边权的异或和,求最大的路径异或和$

分析:

$xor trie配合简单dp思想$
$我们发现这个是个连续异或和,经典的技巧$
$设prefix[i]:=A_1\oplus A_2\oplus\cdots\oplus A_i$
$那么显然我们可以得到xor(l, r)=prefix[r]\oplus prefix[l-1]$
$同理树上也是一样的,把树搜一遍,也就是相当于求f[i]:=以i这个点结尾的路径的最大异或和$
$最终答案ans = max\{f[i]\}$
$时间复杂度O(nb)$

代码:

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//
// Created by TaoSama on 2016-04-22
// Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:102400000,102400000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
struct Trie {
static const int M = 32 * 1e5 + 10, S = 2;
int root, sz;
int nxt[M][S];
int newNode() {
memset(nxt[sz], -1, sizeof nxt[sz]);
return sz++;
}
void init() {
sz = 0;
root = newNode();
}
void insert(int x) {
int u = root;
for(int i = 31; ~i; --i) {
int c = x >> i & 1, &v = nxt[u][c];
if(v == -1) v = newNode();
u = v;
}
}
int query(int x) {
int u = root, ret = 0;
for(int i = 31; ~i; --i) {
int c = x >> i & 1;
if(~nxt[u][c ^ 1]) {
ret |= 1 << i;
u = nxt[u][c ^ 1];
} else u = nxt[u][c];
}
return ret;
}
} trie;
struct Edge {
int v, nxt, c;
} edge[N << 1];
int head[N], cnt;
void addEdge(int u, int v, int c) {
edge[cnt] = {v, head[u], c};
head[u] = cnt++;
edge[cnt] = {u, head[v], c};
head[v] = cnt++;
}
void dfs(int u, int f, int sum, int& ans) {
for(int i = head[u]; ~i; i = edge[i].nxt) {
int v = edge[i].v;
if(v == f) continue;
ans = max(ans, trie.query(sum ^ edge[i].c));
trie.insert(sum ^ edge[i].c);
dfs(v, u, sum ^ edge[i].c, ans);
}
}
int n;
int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
// freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
while(scanf("%d", &n) == 1) {
cnt = 0; memset(head, -1, sizeof head);
for(int i = 1; i < n; ++i) {
int u, v, c; scanf("%d%d%d", &u, &v, &c);
addEdge(u, v, c);
}
int ans = 0;
trie.init(); trie.insert(0);
dfs(0, -1, 0, ans);
printf("%d\n", ans);
}
return 0;
}


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