BZOJ1954 Pku3764 The xor-longest Path

"trie的经典应用" -- by hzwer

我们把每个点到根的xor值记下来，然后找出两个xor值最大的即可（因为(a ^ c) ^ (b ^ c) = a ^ b）

于是用trie把每个数的二进制位记下来，每次query的时候利用贪心，试图走到另一个儿子即可。

 

  1 /**************************************************************
  2     Problem: 1954
  3     User: rausen
  4     Language: C++
  5     Result: Accepted
  6     Time:360 ms
  7     Memory:27368 kb
  8 ****************************************************************/
  9  
 10 #include <cstdio>
 11 #include <algorithm>
 12  
 13 using namespace std;
 14 const int N = 100005;
 15  
 16 struct edge {
 17     int next, to, v;
 18     edge() {}
 19     edge(int _n, int _t, int _v) : next(_n), to(_t), v(_v) {}
 20 } e[N << 1];
 21  
 22 int first[N], tot;
 23  
 24 struct trie_node {
 25     int son[2];
 26 } t[3000005];
 27  
 28 int cnt_trie = 1, root = 1;
 29  
 30 int n, ans;
 31 int v[N], a[40];
 32  
 33 inline int read() {
 34     int x = 0, sgn = 1;
 35     char ch = getchar();
 36     while (ch < '0' || '9' < ch) {
 37         if (ch == '-') sgn = -1;
 38         ch = getchar();
 39     }
 40     while ('0' <= ch && ch <= '9') {
 41         x = x * 10 + ch - '0';
 42         ch = getchar();
 43     }
 44     return sgn * x;
 45 }
 46  
 47 void Add_Edges(int x, int y, int z) {
 48     e[++tot] = edge(first[x], y, z), first[x] = tot;
 49     e[++tot] = edge(first[y], x, z), first[y] = tot;
 50 }
 51  
 52 void dfs(int p, int fa) {
 53     int x, y;
 54     for (x = first[p]; x; x = e[x].next)
 55         if ((y = e[x].to) != fa) {
 56             v[y] = v[p] ^ e[x].v;
 57             dfs(y, p);
 58         }
 59 }
 60  
 61 void insert(int x) {
 62     int now, i, cnt_a = 0;
 63     for (i = 0; i <= 30; ++i)
 64         a[i] = x & 1, x >>= 1;
 65     reverse(a, a + 31);
 66     now = root;
 67     for (i = 0; i <= 30; ++i) {
 68         if (!t[now].son[a[i]]) t[now].son[a[i]] = ++cnt_trie;
 69         now = t[now].son[a[i]];
 70     }
 71 }
 72  
 73 int query(int x) {
 74     int now, i, cnt_a = 0, res = 0;
 75     for (i = 0; i <= 30; ++i)
 76         a[i] = x & 1, x >>= 1;
 77     reverse(a, a + 31);
 78     now = root;
 79     for (i = 0; i <= 30; ++i) {
 80         if (t[now].son[!a[i]]) now = t[now].son[!a[i]], res += (1 << 30 - i);
 81         else now = t[now].son[a[i]];
 82     }
 83     return res;
 84 }
 85  
 86 int main() {
 87     int i, x, y, z; 
 88     n = read();
 89     for (i = 1; i < n; ++i) {
 90         x = read(), y = read(), z = read();
 91         Add_Edges(x, y, z);
 92     }
 93     dfs(1, 0);
 94     for (i = 1; i <= n; ++i)
 95         insert(v[i]);
 96     for (i = 1; i <= n; ++i)
 97         ans = max(ans, query(v[i]));
 98     printf("%d\n", ans);
 99     return 0;
100 }
View Code

 

posted on 2014-12-20 23:00  Xs酱~  阅读(362)  评论(0)  编辑  收藏  举报