POJ 3764 The xor-longest Path
Posted henry-1202
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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 u and 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)
Solution
01Trie经典题,求树上两点路径xor和的最大值。
我们可以把问题转化一下,先dfs一遍求出这棵树的树上xor前缀和,设为(d[i])。
那么两点之间路径的xor和其实就是(d[i])^(d[j])(因为xor的性质是a^a=0)
那么就是01Trie的经典操作了。。求两数的xor最大值。套板子即可。
我怎么又忘记POJ有多组数据啊,我怎么又多组数据又忘记清数组啊。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <set>
#include <map>
#include <vector>
#include <queue>
#define ll long long
#define inf 0x3f3f3f3f
#define il inline
namespace io {
#define in(a) a=read()
#define out(a) write(a)
#define outn(a) out(a),putchar('
')
#define I_int ll
inline I_int read() {
I_int x = 0 , f = 1 ; char c = getchar() ;
while( c < '0' || c > '9' ) { if( c == '-' ) f = -1 ; c = getchar() ; }
while( c >= '0' && c <= '9' ) { x = x * 10 + c - '0' ; c = getchar() ; }
return x * f ;
}
char F[ 200 ] ;
inline void write( I_int x ) {
if( x == 0 ) { putchar( '0' ) ; return ; }
I_int tmp = x > 0 ? x : -x ;
if( x < 0 ) putchar( '-' ) ;
int cnt = 0 ;
while( tmp > 0 ) {
F[ cnt ++ ] = tmp % 10 + '0' ;
tmp /= 10 ;
}
while( cnt > 0 ) putchar( F[ -- cnt ] ) ;
}
#undef I_int
}
using namespace io ;
using namespace std ;
#define N 100100
int n, ch[N * 32][2], tot;
int d[N];
int head[N], cnt;
struct edge {
int to, nxt, v;
}e[N<<1];
void ins(int u, int v, int w) {
e[++cnt] = (edge) {v, head[u], w};
head[u] = cnt;
}
void dfs(int u, int fa) {
for(int i = head[u]; i; i = e[i].nxt) {
int v = e[i].to;
if(e[i].to == fa) continue;
d[v] = d[u] ^ e[i].v;
dfs(e[i].to, u);
}
}
void insert(int x) {
int u = 0;
for(int i = 30; i >= 0; --i) {
int c = (x >> i) & 1;
if(!ch[u][c]) ch[u][c] = ++tot;
u = ch[u][c];
}
}
int query(int x) {
int u = 0, ans = 0;
for(int i = 30; i >= 0; --i) {
int c = (x >> i) & 1;
if(ch[u][c ^ 1]) u = ch[u][c ^ 1], ans += 1ll * (1 << i);
else u = ch[u][c];
}
return ans;
}
int main() {
while(~scanf("%d", &n)) {
memset(head, 0, sizeof(head));
memset(ch, 0, sizeof(ch));
cnt = 0; tot = 0;
for(int i = 1, u, v, w; i < n; ++i) {
scanf("%d%d%d", &u, &v, &w);
ins(u, v, w); ins(v, u, w);
}
memset(d, 0, sizeof(d));
dfs(0, 0);
int ans = 0;
for(int i = 0; i < n; ++i) {
insert(d[i]);
ans = max(ans, query(d[i]));
}
printf("%d
", ans);
}
}
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