bzoj 2115 线性基

这种路径异或问题，可以转换为一条路径和若干个环的线性组合，然后就能用线性基搞了。

复习了一波线性基。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PII pair<int, int>
#define PLI pair<LL, int>
#define ull unsigned long long
using namespace std;
 
const int N = 1e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
 
int n, m;
LL d[N];
bool vis[N];
vector<PLI> edge[N];
struct Base {
    vector<LL> a;
    void add(LL x) {
        for(int i = 0; i < a.size(); i++)
            x = min(x, x^a[i]);
        if(!x) return;
        for(int i = 0; i < a.size(); i++)
            a[i] = min(a[i], a[i]^x);
        a.push_back(x);
    }
    LL getMx(LL ans) {
        for(int i = 0; i < a.size(); i++)
            ans = max(ans, ans^a[i]);
        return ans;
    }
} base;
 
void dfs(int u, int fa) {
    vis[u] = true;
    for(int i = 0; i < edge[u].size(); i++) {
        int v = edge[u][i].se; LL w = edge[u][i].fi;
        if(v == fa) continue;
        if(vis[v]) {
            base.add(d[u]^d[v]^w);
        } else {
            d[v] = d[u] ^ w;
            dfs(v, u);
        }
    }
}
int main() {
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= m; i++) {
        int u, v; LL w;
        scanf("%d%d%lld", &u, &v, &w);
        edge[u].push_back(mk(w, v));
        edge[v].push_back(mk(w, u));
    }
    dfs(1, 0);
    printf("%lld
", base.getMx(d[n]));
    return 0;
}
 
/*
*/