• poj3764 The XOR Longest Path【dfs】【Trie树】


    The xor-longest Path
    Time Limit: 2000MS   Memory Limit: 65536K
    Total Submissions: 10038   Accepted: 2040

    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:

    _{xor}length(p)=oplus_{e in p}w(e)

    ⊕ 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 uand 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)

    Source

    题意:

    给一棵带边权的树,想找得到两个点,他们的路径上的权值异或最小。

    思路:

    首先我们任意找一个作为根,可以用dfs求出其他节点到根的路径的异或,记为xordis

    那么对于树上的任意两个节点i, j,i到j的路径的异或和应该是xordis[i] ^ xordis[j]

    因为i到j的路径,相当于i到根,根到j,其中重叠的部分,他们的异或值正好是0

    因此这道题就变成了找两点异或值最小,https://www.cnblogs.com/wyboooo/p/9824293.html 和这道题就差不多了

    最后还需要注意,search找到的最大值是除根以外的,还需要和xordis比较一下,取较大值。

      1 #include <iostream>
      2 #include <set>
      3 #include <cmath>
      4 #include <stdio.h>
      5 #include <cstring>
      6 #include <algorithm>
      7 #include <map>
      8 using namespace std;
      9 typedef long long LL;
     10 #define inf 0x7f7f7f7f
     11 
     12 int n;
     13 const int maxn = 1e5 + 5;
     14 struct edge{
     15     int v, w;
     16     int nxt;
     17 }e[maxn * 2];
     18 int head[maxn], tot = 0;
     19 int xordis[maxn];
     20 int trie[maxn * 32 + 5][3], treetot = 1;
     21 
     22 void addedge(int u, int v, int w)
     23 {
     24     e[tot].v = v;
     25     e[tot].w = w;
     26     e[tot].nxt = head[u];
     27     head[u] = tot++;
     28     e[tot].v = u;
     29     e[tot].w = w;
     30     e[tot].nxt = head[v];
     31     head[v] = tot++;
     32 }
     33 
     34 void dfs(int rt, int fa)
     35 {
     36     for(int i = head[rt]; i != -1; i = e[i].nxt){
     37         int v = e[i].v;
     38         if(v == fa)continue;
     39         xordis[v] = xordis[rt] ^ e[i].w;
     40         dfs(v, rt);
     41     }
     42 }
     43 
     44 void init()
     45 {
     46     memset(head, -1, sizeof(head));
     47     tot = 0;
     48     memset(xordis, 0, sizeof(xordis));
     49     memset(trie, 0, sizeof(trie));
     50 }
     51 
     52 void insertt(int x)
     53 {
     54     int p = 1;
     55     for(int i = 30; i >= 0; i--){
     56         int ch = x >> i & 1;
     57         if(trie[p][ch] == 0){
     58             trie[p][ch] = ++tot;
     59         }
     60         p = trie[p][ch];
     61     }
     62 }
     63 
     64 int searchh(int x)
     65 {
     66     int p = 1, ans = 0;
     67     for(int i = 30; i >= 0; i--){
     68         int ch = x >> i & 1;
     69         if(trie[p][ch ^ 1]){
     70             p = trie[p][ch ^ 1];
     71             ans |= 1 << i;
     72         }
     73         else{
     74             p = trie[p][ch];
     75         }
     76     }
     77     return ans;
     78 }
     79 
     80 int main()
     81 {
     82     while(scanf("%d", &n) != EOF){
     83         init();
     84         for(int i = 0; i < n - 1; i++){
     85             int u, v, w;
     86             scanf("%d%d%d", &u, &v, &w);
     87             addedge(u, v, w);
     88         }
     89         dfs(0, -1);
     90 
     91         /*for(int i = 0; i < n; i++){
     92             printf("%d
    ", xordis[i]);
     93         }*/
     94 
     95         int ans = 0;
     96         for(int i = 1; i < n; i++){
     97             insertt(xordis[i]);
     98             //cout<<searchh(xordis[i])<<endl;
     99             ans = max(ans, searchh(xordis[i]));
    100             ans = max(ans, xordis[i]);
    101         }
    102         printf("%d
    ", ans);
    103     }
    104     return 0;
    105 }
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  • 原文地址:https://www.cnblogs.com/wyboooo/p/9824427.html
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