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  • POJ 3764 The xor-longest Path (字典树)

    题意:给出一颗n个节点的边权树,求一条路径(u,v),使得路径上的边的权值异或值最大。

    析:先从0开始遍历树,记录所有的点到0的路径的边权异或值,然后任意两点的路径的异或值就是dp[u]^dp[v],

    然后再构造一棵二进制树,每次查询,注意长度要相同,最后求最大值即可。

    代码如下:

    #pragma comment(linker, "/STACK:1024000000,1024000000")
    #include <cstdio>
    #include <string>
    #include <cstdlib>
    #include <cmath>
    #include <iostream>
    #include <cstring>
    #include <set>
    #include <queue>
    #include <algorithm>
    #include <vector>
    #include <map>
    #include <cctype>
    #include <cmath>
    #include <stack>
    #include <sstream>
    #define debug() puts("++++");
    #define gcd(a, b) __gcd(a, b)
    #define lson l,m,rt<<1
    #define rson m+1,r,rt<<1|1
    #define freopenr freopen("in.txt", "r", stdin)
    #define freopenw freopen("out.txt", "w", stdout)
    using namespace std;
    
    typedef long long LL;
    typedef unsigned long long ULL;
    typedef pair<int, int> P;
    const int INF = 0x3f3f3f3f;
    const double inf = 0x3f3f3f3f3f3f;
    const double PI = acos(-1.0);
    const double eps = 1e-8;
    const int maxn = 100000 + 50;
    const LL mod = 2147483648;
    const int dr[] = {-1, 0, 1, 0};
    const int dc[] = {0, 1, 0, -1};
    const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
    int n, m;
    const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    inline bool is_in(int r, int c){
      return r >= 0 && r < n && c >= 0 && c < m;
    }
    struct Edge{
      int to, w, next;
    };
    int head[maxn];
    int cnt;
    Edge a[maxn<<1];
    int dp[maxn];
    
    void add(int u, int v, int w){
      a[cnt].to = v;
      a[cnt].w = w;
      a[cnt].next = head[u];
      head[u] = cnt++;
    }
    bool vis[maxn];
    
    void dfs(int u){
      for(int i = head[u]; ~i; i = a[i].next){
        int v = a[i].to;
        if(vis[v])  continue;
        vis[v] = 1;
        dp[v] = dp[u] ^ a[i].w;
        dfs(v);
      }
    }
    const int maxnode = maxn * 64 + 10;
    const int siga_size = 2;
    
    struct Trie{
      int ch[maxnode][siga_size];
      int val[maxnode];
      int sz;
      void init(){
        sz = 1;
        memset(ch[0], 0, sizeof ch[0]);
      }
    
      void insert(int *s, int v, int len){
        int u = 0;
        for(int i = len; i >= 0; --i){
          int c = s[i];
          if(!ch[u][c]){
            memset(ch[sz], 0, sizeof ch[sz]);
            val[sz] = 0;
            ch[u][c] = sz++;
          }
          u = ch[u][c];
        }
        val[u] = v;
      }
    
      int query(int *s, int len){
        int u = 0;
        for(int i = len; i >= 0; --i){
          int c = s[i]^1;
          if(!ch[u][c])  c ^= 1;
          if(!ch[u][c])  return val[u];
          u = ch[u][c];
        }
        return val[u];
      }
    };
    
    Trie trie;
    
    int main(){
      while(scanf("%d", &n) == 1){
        memset(head, -1, sizeof head);
        cnt = 0;
        for(int i = 1; i < n; ++i){
          int u, v, w;
          scanf("%d %d %d", &u, &v, &w);
          add(u, v, w);
          add(v, u, w);
        }
        memset(vis, 0, sizeof vis);
        vis[0] = 1;  dp[0] = 0;
        dfs(0);
        trie.init();
        for(int i = 0; i < n; ++i){
          int len = -1, x = dp[i];
          int s[50];
          while(x){
            s[++len] = x % 2;
            x /= 2;
          }
          while(len < 30)  s[++len] = 0;
          trie.insert(s, dp[i], len);
        }
        int ans = 0;
        for(int i = 0; i < n; ++i){
          int len = -1, x = dp[i];
          int s[50];
          while(x){
            s[++len] = x % 2;
            x /= 2;
          }
          while(len < 30)  s[++len] = 0;
          ans = max(ans, dp[i]^trie.query(s, len));
        }
        printf("%d
    ", ans);
      }
      return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/dwtfukgv/p/6682467.html
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