CF 724 G. Xor-matic Number of the Graph

G. Xor-matic Number of the Graph

链接

题意：

　　给定一个无向图，一个interesting的三元环(u,v,s)满足，从u到v的路径上的异或和等于s，三元环的权值为s，求所有三元环权值之和。

分析：

　　求出所有的三元环，建立线性基，然后逐位求每一位的贡献。

代码：

#include<cstdio>

#include<algorithm>

#include<cstring>

#include<iostream>

#include<cmath>

#include<cctype>

#include<set>

#include<queue>

#include<vector>

#include<map>

using namespace std;

typedef long long LL;

inline LL read() {

    LL x=,f=;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-;

    for(;isdigit(ch);ch=getchar())x=x*+ch-'';return x*f;

}

const int N = , mod = 1e9 + ;

struct Edge{ int to, nxt; LL w; } e[];

int head[N], En;

LL dis[N], b[], mi[N], Ans;

bool vis[N];

vector<LL> B;

vector<int> A;

inline void add_edge() {

    int u = read(), v = read(); LL w = read();

    ++En; e[En].to = v, e[En].w = w; e[En].nxt = head[u]; head[u] = En;

    ++En; e[En].to = u, e[En].w = w; e[En].nxt = head[v]; head[v] = En;

}

void dfs(int u) {

    A.push_back(u);

    vis[u] = ;

    for (int i = head[u]; i; i = e[i].nxt) {

        int v = e[i].to;

        if (vis[v]) B.push_back(dis[u] ^ e[i].w ^ dis[v]);

        else {

            dis[v] = dis[u] ^ e[i].w;

            dfs(v);

        }

    }

}

void Insert(LL x) {

    for (int i = ; ~i; --i) {

        if ((x >> i) & ) {

            if (b[i]) x ^= b[i];

            else {

                b[i] = x;

                break;

            }

        }

    }

}

void Clear() {

    memset(b, , sizeof(b));

    A.clear(), B.clear();

}

inline void mul(LL &x,LL y) { x *= y; x %= mod; }

inline void add(LL &x,LL y) { x += y; if (x >= mod) x -= mod; }

void Calc() {

    for (int i = ; i < (int)B.size(); ++i) Insert(B[i]);

    LL cnt[] = {, }, r = , now;

    for (int i = ; ~i; --i) if (b[i]) r ++;

    for (int k = ; ~k; --k) {

        bool flag = ; cnt[] = cnt[] = ;

        for (int i = ; ~i; --i) if ((b[i] >> k) & ) flag = ;

        for (int i = ; i < (int)A.size(); ++i) cnt[(dis[A[i]] >> k) & ] ++;

        now = cnt[] * (cnt[] - ) /  + cnt[] * (cnt[] - ) / ; now %= mod;

        if (flag) {

            if (r >= ) mul(now, mi[r - ]);

            mul(now, mi[k]);

            add(Ans, now);

        }

        now = cnt[] * cnt[];

        if (flag) { if (r >= ) mul(now, mi[r - ]); }

        else mul(now, mi[r]);

        mul(now, mi[k]);

        add(Ans, now);

    }

    Clear();

}

int main() {

    int n = read(), m = read();

    mi[] = ;

    for (int i = ; i <= ; ++i) mi[i] = mi[i - ] *  % mod;

    for (int i = ; i <= m; ++i) add_edge();

    for (int i = ; i <= n; ++i) {

        if (!vis[i]) {

            dfs(i);

            Calc();

        }

    }

    cout << Ans;

    return ;

}