标签:字典树 def max inf class 就会 == ima queue
思路(来自题解):
众所周知树上两个点xy的距离是deep[x]+deep[y]-deep[lca(x,y)]*2
然后我们把这个加减法换成异或,我们就会发现,deep[lca(x,y)]被消掉了
所以题目就简化成w是每个点的前缀异或和,只要找到一对最大的(x,y)让w[x]^w[y]最大就行了,这个经典问题用字典树就能解决了。
代码:
#include<set> #include<map> #include<stack> #include<cmath> #include<queue> #include<vector> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> typedef long long ll; const int maxn = 1e5 + 10; const int seed = 131; const ll MOD = 1e9 + 7; const ll INF = 1e17; using namespace std; int tol, node[32 * maxn][2]; ll val[32 * maxn], w[maxn]; void Insert(ll x){ int root = 0; for(int i = 31; i >= 0; i--){ int u = (x >> i) & 1; if(node[root][u] == 0){ memset(node[tol], 0, sizeof(node[tol])); node[root][u] = tol++; } root = node[root][u]; } val[root] = x; } ll query(ll x){ int root = 0; for(int i = 31; i >= 0; i--){ int u = (x >> i) & 1; if(node[root][!u]) root = node[root][!u]; else root = node[root][u]; } return x ^ val[root]; } int main(){ int T, n, a; ll x; scanf("%d", &T); while(T--){ memset(node[0], 0, sizeof(node[0])); tol = 1; w[1] = 0; scanf("%d", &n); for(int i = 2; i <= n; i++){ scanf("%d%lld", &a, &x); w[i] = x ^ w[a]; Insert(w[i]); } ll ans = 0; for(int i = 1; i <= n; i++){ ans = max(ans, query(w[i])); } printf("%lld\n", ans); } return 0; }
FJUT seventh的tired树上路径(01字典树)题解
标签:字典树 def max inf class 就会 == ima queue
原文地址:https://www.cnblogs.com/KirinSB/p/9829321.html