Intel Code Challenge Final Round (Div. 1 + Div. 2, Combined) G

题意：

给定一棵树，每条边有权，定义三元组(u,v,s)为存在一条从u到v的路径，路径xor值为s，且u <= v，问所有这样的三元组的s之和是多少？？

solution：

一条从u到v的路径，总是可以通过任取一条路，xor连通块内的环搞出。随意取出一个连通块处理，先dfs出一棵dfs树，那么一条非树边对应一个环，把环的权值拿去组成一个线性空间，高斯消元求出线性基(今天才发现之前自己写的求基是错的。。。)，假设得到的是r维线性空间

那么对于(u,v,s)，不同的s有2^r种，通过基xor出来就是。对于dfs中的每个节点，处理出它到根的路径的xor

记这些xor值中第i位是0的有ai个，第i位是1的有bi个

对于第k位，如果任意一个基的第k位是1，那么贡献2^k*C(n,2)*2^(r-1)，因为你总能通过基xor出这一位的1

否则贡献是ak*bk*2^k*2^r，因为这时候就要靠原路径xor出来了

#include<iostream>#include<cstdio>#include<queue>#include<vector>#include<bitset>#include<algorithm>#include<cstring>#include<map>#include<stack>#include<set>#include<cmath>#include<ext/pb_ds/priority_queue.hpp>using namespace std;const int maxn = 1E5 + 10;typedef long long LL;LL mo = 1E9 + 7;struct E{int to; LL va;E(){}E(int to,LL va): to(to),va(va){}};int n,m,tot,L[maxn];LL t,f[2][66],ans,w[maxn],A[2*maxn],B[66];bool C[66];vector <E> v[maxn];void Dfs(int x,int fa){++t; LL y = w[x];for (int i = 0; i <= 60; i++,y >>= 1LL)++f[y&1LL][i];for (int i = 0; i < v[x].size(); i++) {int to = v[x][i].to;if (to == fa) continue;if (L[to]) {if (L[x] >= L[to])A[++tot] = (w[x]^w[to]^v[x][i].va);continue;}L[to] = L[x] + 1;w[to] = (w[x]^v[x][i].va);Dfs(to,x);}}bool cmp(const LL &x,const LL &y) {return x > y;}void Solve(int x){memset(f,0,sizeof(f));tot = t = 0;L[x] = 1;Dfs(x,0);memset(B,0,sizeof(B));memset(C,0,sizeof(C));LL cnt = 0;for (LL j = 60; j >= 0; j--)for (int i = 1; i <= tot; i++)if ((A[i] >> j) == 1) {if (!B[j]) B[j] = A[i];else A[i] ^= B[j];}for (int j = 60; j >= 0; j--)if (B[j]) ++cnt;for (int i = 1; i <= tot; i++)for (LL j = 60; j >= 0; j--)if ((A[i] >> j) & 1)C[j] = 1;t = t*(t-1LL)/2LL%mo;LL a = (1LL<<(cnt-1))%mo;LL b = (1LL<<cnt)%mo;LL now = 1;for (int j = 0; j <= 60; j++) {if (C[j]) {ans += t*now%mo*a%mo;ans %= mo;}else {ans += f[0][j]*f[1][j]%mo*now%mo*b%mo;ans %= mo;}now <<= 1LL;now %= mo;}}int main(){#ifdef DMCfreopen("DMC.txt","r",stdin);#endifcin >> n >> m;for (int i = 1; i <= m; i++) {int x,y; LL w;scanf("%d%d%I64d",&x,&y,&w);v[x].push_back(E(y,w));v[y].push_back(E(x,w));}for (int i = 1; i <= n; i++)if (!L[i]) Solve(i);cout << ans;return 0;}