Intel Code Challenge Final Round Xor-matic Number of the Graph -- 线性基

传送门

点x到点y的路径可以是一条从x到y的路径与若干个环的异或和

首先搞出一棵dfs树，并且得到树上每个环的xor值。　　我们发现，两点间就是本来的dis xor 某些环的xor值，即可组合得到一些新的异或值。位运算的题目，我们显然按位来做。　　首先，对于两个这一位同时为1或者同时为0的，我们考虑若要有贡献，必须是从环上得到一个这一位为1的xor值，如果线性基这一位都是0则无贡献，否则我们可以考虑，假设线性基中有r个向量，那么我们把这一位为1的一个向量排除在外，剩下的随便选，任意组合，也就是2r−1，得到一个权值，再根据得到的权值这一位是1还是0，来决定被排除在外的这个向量选不选，所以贡献就是2r−1。　　同理，如果一个是1一个是0，那么我同样是分类讨论，向量中有无这一位是1的，分别算贡献即可。 

///                 .-~~~~~~~~~-._       _.-~~~~~~~~~-.
///             __.'              ~.   .~              `.__
///           .'//                  \./                  \\`.
///        .'//                     |                     \\`.
///       .'// .-~"""""""~~~~-._     |     _,-~~~~"""""""~-. \\`.
///     .'//.-"                 `-.  |  .-'                 "-.\\`.
///   .'//______.============-..   \ | /   ..-============.______\\`.
/// .'______________________________\|/______________________________`.
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <functional>
#include <stdlib.h>
#include <time.h>
#include <bitset>
using namespace std;

#define pi acos(-1)
#define s_1(x) scanf("%d",&x)
#define s_2(x,y) scanf("%d%d",&x,&y)
#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)
#define S_1(x) scan_d(x)
#define S_2(x,y) scan_d(x),scan_d(y)
#define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z)
#define PI acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x,y) memset(x,y,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define fOR(n,x,i) for(int i=n;i>=x;i--)
#define fOr(n,x,i) for(int i=n;i>x;i--)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
#define db double
#define ll long long
#define mp make_pair
#define pb push_back
typedef long long LL;
typedef pair <int, int> ii;
const int INF=0x3f3f3f3f;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
const int dx[]={-1,0,1,0,1,-1,-1,1};
const int dy[]={0,1,0,-1,-1,1,-1,1};
const int maxn=1e5+10;
const int maxx=1e3+10;
const double EPS=1e-8;
const double eps=1e-8;
const int mod=1e9+7;
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
template <class T>
inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}
while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}

inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;
while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}
else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}
if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}
if(IsN) num=-num;return true;}

void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}
void print(LL a){ Out(a),puts("");}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
//cerr << "run time is " << clock() << endl;

LL p[65];
int tot,cnt,r;
int head[maxn];
LL vis[maxn],A[maxn*4];
vector<int>G;
struct edge
{
int v,next;
LL w;
}e[maxn*4];

void add(int u,int v,LL z)
{
e[tot].v=v;
e[tot].w=z;
e[tot].next=head[u];
head[u]=tot++;
}

LL _pow(LL a,LL b)
{
LL ret=1;
W(b)
{
if(b&1) ret=ret*a%mod;
a=a*a%mod;
b>>=1;
}
return ret;
}

void init()
{
me(head,-1);
me(vis,-1);
tot=0;
}

void dfs(int fa,LL s)
{
if(vis[fa]==-1)
{
vis[fa]=s;
}
else
{
A[++cnt]=vis[fa]^s;
return ;
}
G.pb(fa);
for(int i=head[fa];~i;i=e[i].next)
{
int v=e[i].v;
dfs(v,s^e[i].w);
}
}

LL calc()
{
me(p,0);
FOR(1,cnt,i)
{
fOR(63,0,j)
{
if(!(A[i]>>j&1)) continue;
if(!p[j]){p[j]=A[i];break;}
A[i]^=p[j];
}
}
r=0;
fOR(63,0,i)
if(p[i]) r++;//计算线性基有效的向量个数

LL ans=0;
fOR(63,0,i)//按位算贡献
{
int c[2]={0},sign=0;
fOR(63,0,j)
if(p[j]>>i&1) sign=1;//是否存在某个向量的这一位为1
for(int j=0;j<G.size();j++)
{
int u=G[j];
if(sign) ans+=_pow(2,i+r-1)*j%mod;
else ans+=_pow(2,i+r)*c[!(vis[u]>>i&1)]%mod;
ans%=mod;
c[vis[u]>>i&1]++;
}
}
return ans;
}

int n,m;
void solve()
{
W(cin>>n>>m)
{
init();
FOR(1,m,i)
{
int u,v;LL val;
S_3(u,v,val);
add(u,v,val);
add(v,u,val);
}
LL ans=0;
FOR(1,n,i)
{
if(vis[i]==-1)
{
G.clear();
cnt=0;
dfs(i,0);
ans+=calc();
ans%=mod;
}
}
print(ans);
}
}

int main()
{
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
int t=1;
//init();
//s_1(t);
for(int cas=1;cas<=t;cas++)
{
//printf("Case #%d: ",cas);
solve();
}
}