[并查集 杂题] BZOJ 4452 [Cerc2015]Export Estimate

关键就是一个观察

怎么判断是个环 在维护并查集的时候顺便维护下size和度数为2的点的size

#include
#include
#include
using namespace std;

inline char nc(){
  static char buf[100000],*p1=buf,*p2=buf;
  return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*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;


struct edge{
  int u,v,w;
  bool operator < (const edge&B) const{
    return w>B.w;
  }
}ed[N];

int n,m;

int Q,a[N];
int idx[N],nodes[N],edges[N];
inline bool cmp(int x,int y){
  return a[x]>a[y];
}

int num0,num2,numc;
int deg[N];
int fat[N],size[N],size2[N];
inline int Fat(int u){
  return u==fat[u]?u:fat[u]=Fat(fat[u]);
}
inline void del(int u){
  if (deg[u]==0) num0--;
  if (deg[u]==2) num2--,size2[Fat(u)]--;
}
inline void add(int u){
  if (deg[u]==0) num0++;
  if (deg[u]==2) num2++,size2[Fat(u)]++;
}
inline void delc(int u){
  if (size[u]==size2[u]) numc--;
}
inline void addc(int u){
  if (size[u]==size2[u]) numc++;
}


inline void Add(int u,int v){
  int x=Fat(u),y=Fat(v);
  if (x^y){
    delc(x); delc(y); del(u); del(v); 
    deg[u]++; deg[v]++;
    add(u); add(v);
    fat[x]=y,size[y]+=size[x],size2[y]+=size2[x];
    addc(y);
  }else{
    delc(x); del(u); del(v); 
    deg[u]++; deg[v]++;
    add(u); add(v); addc(x);
  }
}

int main(){
  freopen("t.in","r",stdin);
  freopen("t.out","w",stdout);
  read(n); read(m);
  for (int i=1;i<=m;i++)
    read(ed[i].u),read(ed[i].v),read(ed[i].w);
  sort(ed+1,ed+m+1);
  read(Q);
  for (int i=1;i<=Q;i++) read(a[i]),idx[i]=i;
  sort(idx+1,idx+Q+1,cmp);
  for (int i=1;i<=n;i++)
    deg[i]=0,fat[i]=i,size[i]=1,size2[i]=0;
  num0=n;
  int pnt=1;
  int curm=0;
  for (int i=1;i<=Q;i++){
    while (pnt<=m && ed[pnt].w>=a[idx[i]]) Add(ed[pnt].u,ed[pnt].v),curm++,pnt++;
    nodes[idx[i]]=n-num0-num2+numc;
    edges[idx[i]]=curm-num2+numc;
  }
  for (int i=1;i<=Q;i++)
    printf("%d %d\n",nodes[i],edges[i]);
  return 0;
}