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

## 题解

$a\text{ }xor\text{ }b\text{ }xor\text{ }b = a$

## Code

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
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];
G[++cnt].to = v;
G[cnt].dis = d;
}
inline void insert(int u,int v,int d){
}
int dis[maxn];
#define v G[i].to
void dfs(int u,int fa){
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(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){
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;
}


posted @ 2017-02-19 21:38  Sky_miner  阅读(143)  评论(0编辑  收藏  举报