[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<cstdio>#include<cstdlib>#include<algorithm>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;}