[伪位运算生成树] HDU 5629 Clarke and tree

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求边权and和最大生成树

从高位到低位枚举看这位为1的以及满足当前答案（答案上为1的位则边权这一位也要为1）的边是否能构成一棵生成树

#include<cstdio>#include<cstdlib>#include<algorithm>using namespace std;inline char nc(){  static char buf[100000],*p1=buf,*p2=buf;  if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }  return *p1++;}inline void read(int &x){  char c=nc(),b=1;  for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;  for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;}const int N=300005;int fat[N];inline void init(int n){  for (int i=1;i<=n;i++) fat[i]=i;}inline int Fat(int u){  return u==fat[u]?u:fat[u]=Fat(fat[u]);}inline bool Merge(int x,int y){  x=Fat(x); y=Fat(y); if (x==y) return 0;  fat[x]=y; return 1;}int n,m,Ans;int u[N],v[N],w[N];int vst[N];int main(){  int Q;  freopen("t.in","r",stdin);  freopen("t.out","w",stdout);  read(Q);  while (Q--){    Ans=0;    read(n); read(m);    for (int i=1;i<=m;i++)      read(u[i]),read(v[i]),read(w[i]),vst[i]=1;    for (int k=31;~k;k--){      init(n); int tem=n;      for (int i=1;i<=m;i++)if (vst[i] && (w[i]>>k&1) && Merge(u[i],v[i]))  tem--;      if (tem==1){Ans+=1<<k;for (int i=1;i<=m;i++)  if (!(w[i]>>k&1))    vst[i]=0;      }    }    printf("%d\n",Ans);  }  return 0;}

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