http://acm.hdu.edu.cn/showproblem.php?pid=4757
函数式trie,和cot差不多,用父节点递推建立子节点,此题学到了一个函数式的小技巧,就是一开始(即root[0])并不需要把所有的节点都建出来,只要让root[0]所有的后继都指向自己就可以了。
#include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <queue> #include <algorithm> #include <vector> #include <cstring> #include <stack> #include <cctype> #include <utility> #include <map> #include <string> #include <climits> #include <set> #include <string> #include <sstream> #include <utility> #include <ctime> #include <bitset> #pragma comment(linker, "/STACK:102400000,102400000") using std::priority_queue; using std::vector; using std::swap; using std::stack; using std::sort; using std::max; using std::min; using std::pair; using std::map; using std::string; using std::cin; using std::cout; using std::set; using std::queue; using std::string; using std::stringstream; using std::make_pair; using std::getline; using std::greater; using std::endl; using std::multimap; using std::deque; using std::unique; using std::lower_bound; using std::random_shuffle; using std::bitset; using std::upper_bound; using std::multiset; typedef long long LL; typedef unsigned long long ULL; typedef unsigned UN; typedef pair<int, int> PAIR; typedef multimap<int, int> MMAP; typedef LL TY; typedef long double LF; const int MAXN(4000010); const int MAXM(50010); const int MAXE(150010); const int MAXK(6); const int HSIZE(13131); const int SIGMA_SIZE(4); const int MAXH(20); const int INFI((INT_MAX-1) >> 1); const ULL BASE(31); const LL LIM(1e13); const int INV(-10000); const int MOD(31313); const double EPS(1e-7); const LF PI(acos(-1.0)); template<typename T> inline void checkmax(T &a, T b){if(b > a) a = b;} template<typename T> inline void checkmin(T &a, T b){if(b < a) a = b;} template<typename T> inline T ABS(const T &a){return a < 0? -a: a;} struct EDGE { int v, next; } edge[200010]; int first[100010], rear; void init(int n) { rear = 0; memset(first, -1, sizeof(first[0])*(n+1)); } void insert(int u, int v) { edge[rear].v = v; edge[rear].next = first[u]; first[u] = rear++; } struct LCA { int dep[200010], E[200010], pos[100010], fa[100010], table[19][200010]; int back; inline int cmp(int a, int b){ return dep[a] < dep[b]? a: b;} void dfs(int u, int d, int f) { fa[u] = f; E[++back] = u; dep[back] = d; pos[u] = back; for(int i = first[u]; ~i; i = edge[i].next) if(edge[i].v != f) { dfs(edge[i].v, d+1, u); E[++back] = u; dep[back] = d; } } void init() { back = 0; dfs(1, 0, 0); for(int i = 1; i <= back; ++i) table[0][i] = i; for(int i = 1; (1 << i) <= back; ++i) for(int j = 1; j+(1 << i)-1 <= back; ++j) table[i][j] = cmp(table[i-1][j], table[i-1][j+(1 << (i-1))]); } int query(int a, int b) { a = pos[a]; b = pos[b]; if(a > b) swap(a, b); int len = b-a+1, temp = 0; while((1 << temp) <= len) ++temp; --temp; return E[cmp(table[temp][a], table[temp][b-(1 << temp)+1])]; } } lca; struct NODE { int sum; NODE *ch[2]; } pool[MAXN], *root[100010]; int back; NODE *newnode() { pool[back].ch[0] = pool[back].ch[1] = 0; pool[back].sum = 0; return pool+back++; } void updata(int dep, int val, NODE *prt, NODE *&rt) { rt = pool+back++; *rt = *prt; ++rt->sum; if(dep == -1) return; int id = (val >> dep)&1; updata(dep-1, val, prt->ch[id], rt->ch[id]); } int query(int val, NODE *ffrt, NODE *frt, NODE *lrt, NODE *rrt) { int temp = 0; for(int i = 15; i >= 0; --i) { int id = (val >> i)&1; int sm = lrt->ch[id^1]->sum+ rrt->ch[id^1]->sum- frt->ch[id^1]->sum- ffrt->ch[id^1]->sum; if(sm) id^= 1; temp = (temp << 1)|id; lrt = lrt->ch[id]; rrt = rrt->ch[id]; frt = frt->ch[id]; ffrt = ffrt->ch[id]; } return temp^val; } int arr[100010]; void dfs(int u, int f) { updata(15, arr[u], root[f], root[u]); for(int i = first[u]; ~i; i = edge[i].next) if(edge[i].v != f) dfs(edge[i].v, u); } int main() { int n, m; while(~scanf("%d%d", &n, &m)) { back = 0; root[0] = newnode(); root[0]->ch[0] = root[0]->ch[1] = root[0]; for(int i = 1; i <= n; ++i) scanf("%d", arr+i); int u, v; init(n); for(int i = 1; i < n; ++i) { scanf("%d%d", &u, &v); insert(u, v); insert(v, u); } lca.init(); dfs(1, 0); int val; for(int i = 0; i < m; ++i) { scanf("%d%d%d", &u, &v, &val); printf("%d\n", query(val, root[lca.fa[lca.query(u, v)]], root[lca.query(u, v)], root[u], root[v])); } } return 0; }