【题意】给出一棵树N(<=10^5)结点,f(u,v)表示u-v路径上所有边值的异或值(u可以等于v,值为0),Q(<=10)个询问,每个询问x,求出所有f(u,v)点对值为x的数量;
【分析】由于异或性质:同一个数异或两次后等于没有异或;则f(u,v) = f(1,u)^f(1,v);这样dfs一遍求出所有点到1的异或值;然后每次询问O(n),由于x^y=s可以转换为x^s = y;而且最大数字只有10^5;任意两个异或值最大为131071;开个数组hash即可;
【AC CODE】530ms
#include <bits/stdc++.h> using namespace std; typedef long long LL; #define rep(i,a,n) for(int i = a; i < n; i++) #define repe(i,a,n) for(int i = a; i <= n; i++) #define per(i,n,a) for(int i = n; i >= a; i--) #define clc(a,b) memset(a,b,sizeof(a)) const int INF = 0x3f3f3f3f, MAXN = (1e5)+10,MAXIN = 1e4,MAXOUT = 1e4; char buf[MAXIN], *ps = buf, *pe = buf+1; inline void rnext(){ if(++ps == pe) pe = (ps = buf)+fread(buf,sizeof(char),sizeof(buf)/sizeof(char),stdin); } template <class T> inline bool in(T &ans) { ans = 0; T f = 1; if(ps == pe) return false; do{ rnext(); if('-' == *ps) f = -1;} while(!isdigit(*ps) && ps != pe); if(ps == pe) return false; do{ ans = (ans<<1)+(ans<<3)+*ps-48;rnext();}while(isdigit(*ps) && ps != pe); ans *= f; return true; } char bufout[MAXOUT], outtmp[50],*pout = bufout, *pend = bufout+MAXOUT; inline void write(){ fwrite(bufout,sizeof(char),pout-bufout,stdout);pout = bufout;} inline void out_char(char c){ *(pout++) = c;if(pout == pend) write();} inline void out_str(char *s) { while(*s){ *(pout++) = *(s++); if(pout == pend) write(); } } template <class T> inline void out_int(T x) { if(!x){ out_char('0');return;} if(x < 0) x = -x,out_char('-'); int len = 0; while(x){ outtmp[len++] = x%10+48; x /= 10;} outtmp[len] = 0; for(int i = 0, j = len-1; i < j; i++,j--) swap(outtmp[i],outtmp[j]); out_str(outtmp); } int head[MAXN],tol,nxt[MAXN<<1],to[MAXN<<1],cost[MAXN<<1]; inline void add_edge(int u, int v, int c) { nxt[tol] = head[u],to[tol] = v,cost[tol] = c; head[u] = tol++; } int x[MAXN],cnt[131071+10]; void dfs(int u, int fa) { for(int i = head[u]; ~i; i = nxt[i]) { int v = to[i]; if(v == fa) continue; x[v] = x[u]^cost[i]; dfs(v,u); } } int main() { #ifdef SHY freopen("d:\\1.txt", "r", stdin); #endif int t; in(t); while(t--) { int n; in(n); clc(head,-1);tol = 0; rep(i,1,n) { int u,v,c; in(u);in(v);in(c); add_edge(u,v,c); add_edge(v,u,c); } dfs(1,-1); int q; in(q); while(q--) { int s; LL ans = 0; in(s); clc(cnt,0); repe(i,1,n) { cnt[x[i]]++; ans += cnt[x[i]^s]; } out_int(ans),out_char('\n'); } } write(); return 0; }