给你一棵N个点的树，每个点都有权值，然后给N-1条树上的边，之后有Q次询问，问u到v的这条链上的点，和给出权值z异或的最大值。
那么，就是可持久化字典树，然后再加上一个LCA即可求解该问题了，多组输入。利用树上差分，求链信息。
#include #include #include #include #include #include #include #include #include #include #include #include #include //#include //#include #define lowbit(x) ( x&(-x) ) #define pi 3.141592653589793 #define e 2.718281828459045 #define eps 1e-8 #define INF 0x3f3f3f3f #define HalF (l + r)>>1 #define lsn rt<<1 #define rsn rt<<1|1 #define Lson lsn, l, mid #define Rson rsn, mid+1, r #define QL Lson, ql, qr #define QR Rson, ql, qr #define myself rt, l, r #define MP(x, y) make_pair(x, y) using namespace std; typedef unsigned long long ull; typedef unsigned int uit; typedef long long ll; const int maxN = 1e5 + 7; int N, Q, head[maxN], cnt, root[maxN], tot, a[maxN]; struct Eddge { int nex, to; Eddge(int a=-1, int b=0):nex(a), to(b) {} } edge[maxN << 1]; inline void addEddge(int u, int v) { edge[cnt] = Eddge(head[u], v); head[u] = cnt++; } inline void _add(int u, int v) { addEddge(u, v); addEddge(v, u); } struct Trie { int nex[2], s; Trie() { nex[0] = nex[1] = 0; s = 0; } inline void clear() { nex[0] = nex[1] = 0; s = 0; } } t[maxN * 20]; void Insert(int u, int old, int val) { root[u] = ++tot; int rt = root[u]; for(int i=15, id; i>=0; i--) { id = (val >> i) & 1; t[rt] = t[old]; t[rt].s ++; t[++tot].clear(); t[rt].nex[id] = tot; old = t[old].nex[id]; rt = tot; } t[rt] = t[old]; t[rt].s ++; } inline int query(int Lrt, int Rrt, int val) { int ans = 0; for(int i=15, id; i>=0; i--) { id = (val >> i) & 1; if(t[t[Rrt].nex[id ^ 1]].s > t[t[Lrt].nex[id ^ 1]].s) { ans |= 1 << i; Rrt = t[Rrt].nex[id ^ 1]; Lrt = t[Lrt].nex[id ^ 1]; } else { Rrt = t[Rrt].nex[id]; Lrt = t[Lrt].nex[id]; } } return ans; } int fa[maxN][20], LOG_2[maxN], deep[maxN]; void dfs(int u, int p) { fa[u][0] = p; deep[u] = deep[p] + 1; for(int i=0; (1 << (i + 1)) < N; i++) fa[u][i + 1] = fa[fa[u][i]][i]; Insert(u, root[p], a[u]); for(int i=head[u], v; ~i; i=edge[i].nex) { v = edge[i].to; if(v == p) continue; dfs(v, u); } } inline int _LCA(int u, int v) { if(deep[u] < deep[v]) swap(u, v); int det = deep[u] - deep[v]; for(int i=LOG_2[det]; i>=0; i--) { if((det >> i) & 1) u = fa[u][i]; } if(u == v) return u; for(int i=LOG_2[deep[u]]; i>=0; i--) { if(fa[u][i] ^ fa[v][i]) { u = fa[u][i]; v = fa[v][i]; } } return fa[u][0]; } inline void init() { cnt = tot = 0; for(int i=1; i<=N; i++) head[i] = -1; for(int i = 1, j = 2, k = 0; i <= N; i++) { if(i == j) { j <<= 1; k++; } LOG_2[i] = k; } } int main() { while(scanf("%d%d", &N, &Q) != EOF) { init(); for(int i=1; i<=N; i++) scanf("%d", &a[i]); for(int i=1, u, v; i
