【JZOJ6357】小ω的图(graph)

description

---

analysis

• 拆位从高位到低位贪心

• 对于当前位，如果把所有当前位为(1)的边塞入，(1)和(n)连通，则该位必须为(1)

• 这个是因为高位的(1)比所有低位的(1)都要优，用并查集维护连通性

• 对固定下的位，继续向下贪心，找低位中满足所有条件的(1)位即可

---

code

#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXN 100005
#define MAXM 500005
#define ll long long
#define reg register ll
#define fo(i,a,b) for (reg i=a;i<=b;++i)
#define fd(i,a,b) for (reg i=a;i>=b;--i)

using namespace std;

ll x[MAXM],y[MAXM],z[MAXM];
ll fa[MAXN],pow[70],f[70];
ll n,m,ans;

inline ll read()
{
	ll x=0,f=1;char ch=getchar();
	while (ch<'0' || '9'<ch){if (ch=='-')f=-1;ch=getchar();}
	while ('0'<=ch && ch<='9')x=x*10+ch-'0',ch=getchar();
	return x*f;
}
inline ll getfa(ll x){return !fa[x]?x:fa[x]=getfa(fa[x]);}
inline void link(ll x,ll y){if (getfa(x)!=getfa(y))fa[getfa(x)]=getfa(y);}
int main()
{
	//freopen("T1.in","r",stdin);
	freopen("graph.in","r",stdin);
	freopen("graph.out","w",stdout);
	n=read(),m=read(),pow[0]=1;
	fo(i,1,62)pow[i]=pow[i-1]*2;
	fo(i,1,m)x[i]=read(),y[i]=read(),z[i]=read();
	fd(j,62,0)
	{
		ans+=pow[j],memset(fa,0,sizeof(fa));
		fo(i,1,m)if ((ans&z[i])==ans)link(x[i],y[i]);
		if (getfa(1)!=getfa(n))ans-=pow[j];
	}
	printf("%lld
",ans);
	return 0;
}