首页 > 代码库 > POJ 3764 The xor-longest Path (字典树)
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\n", ans); } return 0; }
POJ 3764 The xor-longest Path (字典树)
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