bzoj2322 梦想封印

题意和题解见思路索引。

标程及易错点：

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 ll read()
 {
     ll x=;char ch=getchar();
     while (ch<''||ch>'') ch=getchar();
     while (''<=ch&&ch<='') x=(x<<)+(x<<)+ch-'',ch=getchar();
     return x;
 }
 const int M=;
 const int N=;
 set<ll> s;
 int cnt,head[N],blo,u[M],v[M],top,x,y,pos,vis[N],tag[M*],n,m,Q,e[M];
 ll ans[M],ww,w[N],base[],q[N];
 set<ll>::iterator it;
 struct node{int to,next;ll w;}num[M*];
 void add(int x,int y,ll w)
 {num[++cnt].to=y;num[cnt].next=head[x];num[cnt].w=w;head[x]=cnt;}
 void ins(ll x)
 {
     if (!x) return;
     for (int i=;i>=;i--)//从大到小
       if ((x>>i)&) {base[i]=x;blo++;break;}
     for (top=,it=s.begin();it!=s.end();++it) q[++top]=min(*it,*it^x);
     s.clear();
     for (int j=;j<=top;j++) s.insert(q[j]);
 }
 ll match(ll x)
 {
     for (int i=;i>=&&x;i--)//从高位到低位消
       if ((x^base[i])<x) x^=base[i];//能消的位都消元
     return x;
 }
 void dfs(int x,int fa)
 {
    vis[x]=;s.insert(match(w[x]));
    for (int i=head[x];i;i=num[i].next)
      if (num[i].to!=fa&&!tag[i])
      {
         if (!vis[num[i].to]) w[num[i].to]=w[x]^num[i].w,dfs(num[i].to,x);
         else ins(match(w[x]^w[num[i].to]^num[i].w));//这里加环也要对set消元
      }
 }
 int main()
 {
     n=read();m=read();Q=read();
     for (int i=;i<=m;i++) u[i]=read(),v[i]=read(),ww=read(),add(u[i],v[i],ww),add(v[i],u[i],ww);
     for (int i=;i<=Q;i++) e[i]=read(),tag[e[i]*-]=tag[e[i]*]=;
     dfs(,-);
     ans[Q+]=(ll)s.size()*(1ll<<blo)-;
     for (int i=Q;i>=;i--)
     {
         tag[e[i]*-]=tag[e[i]*]=;
         x=u[e[i]],y=v[e[i]],ww=num[e[i]*].w;
         if (vis[x]&&vis[y]) ins(match(w[x]^w[y]^ww));
         else if (vis[x]) w[y]=w[x]^ww,dfs(y,x);
         else if (vis[y]) w[x]=w[y]^ww,dfs(x,y);
         ans[i]=(ll)s.size()*(1ll<<blo)-;
     }
     for (int i=;i<=Q+;i++) printf("%lld\n",ans[i]);
     return ;
 }