bzoj 1954 & poj 3764 The xor-longest Path dfs+Trie

题目大意

给定一棵n个点的带权树,求树上最长的异或和路径

题解

因为\(xor\)操作满足可结合性,所以有
\(a\text{ }xor\text{ }b\text{ }xor\text{ }b = a\)
那么我们可以计算出每个点到根的xor距离,设为\(dis\)
那么我们知道\(dis_u\text{ }xor\text{ }dis_v\)即\(u,v\)之间的距离的xor值
所以我们把所有的\(dis\)插到01Trie里,再对每个\(dis\)值求最大即可

Code

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
inline void read(int &x){
	x=0;char ch;bool flag = false;
	while(ch=getchar(),ch<'!');if(ch == '-') ch=getchar(),flag = true;
	while(x=10*x+ch-'0',ch=getchar(),ch>'!');if(flag) x=-x;
}
inline int cat_max(const int &a,const int &b){return a>b ? a:b;}
inline int cat_min(const int &a,const int &b){return a<b ? a:b;}
const int maxn = 100010;
struct Edge{
	int to,next,dis;
}G[maxn<<1];
int head[maxn],cnt;
void add(int u,int v,int d){
	G[++cnt].to = v;
	G[cnt].next = head[u];
	head[u] = cnt;
	G[cnt].dis = d;
}
inline void insert(int u,int v,int d){
	add(u,v,d);add(v,u,d);
}
int dis[maxn];
#define v G[i].to
void dfs(int u,int fa){
	for(int i = head[u];i;i=G[i].next){
		if(v == fa) continue;
		dis[v] = dis[u]^G[i].dis;
		dfs(v,u);
	}
}
#undef v
int ch[maxn*32][2],nodecnt;
bool ed[maxn*32];
inline void insert(int x){
	int nw = 0;
	for(int i = 31;i;--i){
		int id = (bool)(x & (1 << (i-1)));
		if(ch[nw][id] == 0) ch[nw][id] = ++nodecnt;
		nw = ch[nw][id];
	}ed[nw] = true;
}
inline int query(int x){
	int ret = 0,nw = 0;
	for(int i=31;i;--i){
		int id = (bool)(x & (1 << (i-1)));
		if(ch[nw][id^1] != 0){
			ret |= (1<<(i-1));
			nw = ch[nw][id^1];
		}else nw = ch[nw][id];
	}return ret;
}
inline void init(){
	memset(head,0,sizeof head);
	memset(ch,0,sizeof ch);
	memset(ed,0,sizeof ed);
	memset(dis,0,sizeof dis);
	cnt = nodecnt = 0;
}
int main(){
	int n;
	while(scanf("%d",&n) != EOF){
		init();
		int u,v,d;
		for(int i=1;i<n;++i){
			read(u);read(v);read(d);
			insert(u,v,d);
		}dfs(1,0);
		for(int i=1;i<=n;++i) insert(dis[i]);
		int ans = 0;
		for(int i=1;i<=n;++i){
			ans = max(ans,query(dis[i]));
		}printf("%d\n",ans);		
	}
	getchar();getchar();
	return 0;
}