回归第七题

直接dfs找环即可
两次异或可以抵消，所以直接记录每次异或，最后找到环判断xor[u]^xor[v]dis<u,v>是否为零即可

#include<cstdio>
#include<map>
using namespace std;
const int N = 100000 + 10 , M =  200000 + 10 ;
struct edge
{
    int u , v , nxt ;
    int w ;
}e[M * 2];
int head[M * 2] , E = 0 ;
bool vis [N] ;
int sumxor[N] ;
int n , m ;
int u , v , w ;
bool flag ;

void add (int u , int v , int w)
{
    e[E].u = u , e[E].v = v , e[E].w = w , e[E].nxt = head[u] ;
    head[u] = E ++ ;
    e[E].u = v , e[E].v = u , e[E].w = w , e[E].nxt = head[v] ;
    head[v] = E ++ ;
}

void dfs (int u)
{
    for (int i = head[u] ; i != -1 ; i = e[i].nxt) {
        if (! vis[e[i].v]) {
            vis[e[i].v] = 1 ;
            sumxor[e[i].v] = sumxor[u] ^ e[i].w ;
            dfs (e[i].v) ;
        }
        else {
            if ( (sumxor[e[i].v] ^ sumxor[u] ^ e[i].w) ) {
                flag = 1 ;
                return ;
            }
        }
    }
}

int main ()
{
    //freopen ("a.txt" , "r" , stdin ) ;
    while (~ scanf ("%d%d" , &n , &m) ) {
        E = 0 ;
        flag = 0 ;
        fill (head , head + n + 1 , -1 ) ;
        while (m --) {
            scanf ("%d%d%d" , &u , &v , &w) ;
            add (u , v , w) ;
        }
        fill (vis , vis + n + 1 , 0) ;
        vis[1] = 1 ;
        sumxor[1] = 0 ;
        dfs (1) ;
        printf ("%s\n" , flag ? "Yes" : "No" ) ;
    }
    return 0 ;
}