ACwing.228 异或(线性基)

题目：

• 给一张图，让你找一条1到n的路径，可以重复走边和点，问路径权值异或和最大是多少。

题解：

• 基本思路就是对每条路径把路径上的环的权值异或和放进线性基，因为环要走就走全部一次，走两次会异或掉。所以我们dfs找环，把环放进一个线性基，最后把1到n的简单路径异或和放进去，如果1到n有多个路径，就有环dfs就放进线性基中了，所以求最大值就好了。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<map>
#include<queue>
#include<vector>
#include<string>
#include<fstream>


using namespace std;
typedef long long ll;
const int N = 5e4 + 105;
const int mod = 10000;
const int P = 9999;
const double Pi = acos(- 1.0);
const int INF = 0x3f3f3f3f;


inline int read() {
	char ch = getchar(); int x = 0, f = 1;
	while (ch < '0' || ch > '9') {
		if (ch == '-') f = -1;
		ch = getchar();
	}
	while ('0' <= ch && ch <= '9') {
		x = x * 10 + ch - '0';
		ch = getchar(); 
	} return x * f;
}

//线性基
const int maxbit = 62;		//maxbit不能太大

struct L_B{
	ll lba[maxbit];
	int cnt;
	L_B(){
		memset(lba, 0, sizeof(lba));
		cnt = 0;
	}
	
	void Insert(ll val){		//插入
		for(int i = maxbit - 1; i >= 0; -- i)
			if(val & (1ll << i)){
				if(!lba[i]){
					lba[i] = val;
					break;
				}
				val ^= lba[i];
			}
	}
	

	ll query_max(ll x){		//查找最大值
		ll res = x;
		for(int i = maxbit - 1; i >= 0; -- i){
		    ll che = res ^ lba[i];
			if(che > res) res^=lba[i];
		}
		return res;
	}
};

//图论基本模板

struct Edge{
	int to;
	ll c;
	int nxt;
}edge[N * 4];

int head[N], cnt = 0;
int n, m;

inline void add(int u, int v, ll w){
	edge[cnt].to = v;
	edge[cnt].c = w;
	edge[cnt].nxt = head[u];
	head[u] = cnt ++;
}

//main()
ll res;
int vis[N];
L_B lb;
ll dis[N];

void dfs(int u, int pre){
	vis[u] = 1;
	for(int i = head[u]; i != -1; i = edge[i].nxt){
		int v = edge[i].to;
		ll w = edge[i].c;
		if(v == pre) continue;
		if(vis[v]){
			lb.Insert(w ^ dis[u] ^ dis[v]);
			continue;
		}
		dis[v] = dis[u] ^ w;
		dfs(v, u);
	}
}

int main()
{
	scanf("%d%d",&n,&m);
	memset(head, -1, sizeof(head));
	for(int i = 1; i <= m; ++ i){
		int x, y;
		ll w;
		scanf("%d%d%lld",&x,&y,&w);
		add(x, y, w);
		add(y, x, w);
	}
	dfs(1, -1);
	printf("%lld
",lb.query_max(dis[n]));
	return 0;
}