time limit per test 3 seconds
memory limit per test 512 megabytes
input standard input
output standard output
You are given an undirected graph with weighted edges. The length of some path between two vertices is the bitwise xor of weights of all edges belonging to this path (if some edge is traversed more than once, then it is included in bitwise xor the same number of times). You have to find the minimum length of path between vertex 1 and vertex n.
Note that graph can contain multiple edges and loops. It is guaranteed that the graph is connected.
Input
The first line contains two numbers n and m (1 ≤ n ≤ 100000, n - 1 ≤ m ≤ 100000) — the number of vertices and the number of edges, respectively.
Then m lines follow, each line containing three integer numbers x, y and w (1 ≤ x, y ≤ n, 0 ≤ w ≤ 108). These numbers denote an edge that connects vertices x and y and has weight w.
Output
Print one number — the minimum length of path between vertices 1 and n.
Examples
Input
3 3
1 2 3
1 3 2
3 2 0
Output
2
Input
2 2
1 1 3
1 2 3
Output
0
【翻译】求1 to n的异或最短路。
题解：
①线基入门题。啊，终于入门了。
②最短路转树维护环这个思路很好~
#include<stdio.h>
#define go(i,a,b) for(int i=a;i<=b;i++)
#define ro(i,a,b) for(int i=a;i>=b;i--)
#define fo(i,a,x) for(int i=a[x],v=e[i].v;i;i=e[i].next,v=e[i].v)
const int N=100003;
struct E{int v,next,w;}e[N<<1];
int n,m,U,V,W,head[N],k=1,d[N];bool vis[N];
void ADD(int u,int v,int w){e[k]=(E){v,head[u],w};head[u]=k++;}
struct Linear_Base
{
int a[40];
void Insert(int x){if(x)ro(i,30,0)if((1<<i)&x)!a[i]?a[i]=x:i=0,x^=a[i];}
int Get_Max(int x){ro(i,30,0)if(x>(x^a[i]))x^=a[i];return x;}
}Base;
void dfs(int u)
{
fo(i,head,u)!vis[v]?d[v]=d[u]^e[i].w,vis[v]=1,
dfs(v):Base.Insert(d[u]^d[v]^e[i].w);
}
int main()
{
freopen("in.in","r",stdin);
scanf("%d%d",&n,&m);d[1]=0;
go(i,1,m)scanf("%d%d%d",&U,&V,&W),ADD(U,V,W),ADD(V,U,W);
dfs(1);printf("%d
",Base.Get_Max(d[n]));return 0;
}//Paul_Guderian
这就你，完美的生命,悲哀而真实。——————汪峰《哭泣的拳头》