BZOJ3514 : Codechef MARCH14 GERALD07加強版

Claris發表於2014-05-09

以邊編號為權值

用Link-cut Tree維護最大生成樹

對於新加的第i條邊(u,v)

a[i]表示當a[i]這條邊加入後連通塊個數會減少

若u==v則a[i]=m

若u與v不連通則連上,a[i]=0

若u與v連通則a[i]為u,v路徑上最小值,將那條邊斷開,連上這條邊

查詢[l,r]等價於查詢[l,r]裡有多少a[i]<l

主席樹維護

時間複雜度$O((m+k)\log m)$

 

#include<cstdio>
#define N 400010
#define E 200010
#define M 4000010
int f[N],son[N][2],val[N],min[N],tmp[N];bool rev[N];
inline void read(int&a){char c;while(!(((c=getchar())>='0')&&(c<='9')));a=c-'0';while(((c=getchar())>='0')&&(c<='9'))(a*=10)+=c-'0';}
inline void swap(int&a,int&b){int c=a;a=b;b=c;}
inline int Min(int a,int b){return a<b?a:b;}
inline bool isroot(int x){return !f[x]||son[f[x]][0]!=x&&son[f[x]][1]!=x;}
inline void rev1(int x){if(!x)return;swap(son[x][0],son[x][1]);rev[x]^=1;}
inline void pb(int x){if(rev[x])rev1(son[x][0]),rev1(son[x][1]),rev[x]=0;}
inline void up(int x){min[x]=Min(val[x],Min(min[son[x][0]],min[son[x][1]]));}
inline void rotate(int x){
  int y=f[x],w=son[y][1]==x;
  son[y][w]=son[x][w^1];
  if(son[x][w^1])f[son[x][w^1]]=y;
  if(f[y]){
    int z=f[y];
    if(son[z][0]==y)son[z][0]=x;
    if(son[z][1]==y)son[z][1]=x;
  }
  f[x]=f[y];f[y]=x;son[x][w^1]=y;up(y);
}
inline void splay(int x){
  int s=1,i=x,y;tmp[1]=i;
  while(!isroot(i))tmp[++s]=i=f[i];
  while(s)pb(tmp[s--]);
  while(!isroot(x)){
    y=f[x];
    if(!isroot(y)){if((son[f[y]][0]==y)^(son[y][0]==x))rotate(x);else rotate(y);}
    rotate(x);
  }
  up(x);
}
inline void access(int x){for(int y=0;x;y=x,x=f[x])splay(x),son[x][1]=y,up(x);}
inline int root(int x){access(x);splay(x);while(son[x][0])x=son[x][0];return x;}
inline void makeroot(int x){access(x);splay(x);rev1(x);}
inline void link(int x,int y){makeroot(x);f[x]=y;access(x);}
inline void cutf(int x){access(x);splay(x);f[son[x][0]]=0;son[x][0]=0;up(x);}
inline void cut(int x,int y){makeroot(x);cutf(y);}
inline int ask(int x,int y){makeroot(x);access(y);splay(y);return min[y];}
int n,m,i,k,type,x,y,u[E],v[E],a[E],tot,l[M],r[M],sum[M],head[E],last;
int ins(int x,int a,int b,int c){
  int y=++tot;
  sum[y]=sum[x]+1;
  if(a==b)return y;
  int mid=(a+b)>>1;
  if(c<=mid)l[y]=ins(l[x],a,mid,c),r[y]=r[x];else l[y]=l[x],r[y]=ins(r[x],mid+1,b,c);
  return y;
}
int query(int x,int a,int b,int c){
  if(!x)return 0;
  if(b<=c)return sum[x];
  int mid=(a+b)>>1,t=query(l[x],a,mid,c);
  if(c>mid)t+=query(r[x],mid+1,b,c);
  return t;
}
int main(){
  read(n);read(m);read(k);read(type);
  for(i=0;i<=n;i++)val[i]=min[i]=1000000000;
  for(i=1;i<=m;i++)val[i+n]=min[i+n]=i;
  for(i=1;i<=m;i++){
    read(u[i]);read(v[i]);
    if(u[i]==v[i])a[i]=m;
    else if(root(u[i])==root(v[i]))a[i]=ask(u[i],v[i]),cut(u[a[i]],a[i]+n),cut(v[a[i]],a[i]+n),link(u[i],i+n),link(v[i],i+n);
    else link(u[i],i+n),link(v[i],i+n);
  }
  for(i=1;i<=m;i++)head[i]=ins(head[i-1],0,m,a[i]);
  while(k--){
    read(x),read(y);
    if(type)x^=last,y^=last;
    printf("%d\n",last=n-query(head[y],0,m,x-1)+query(head[x-1],0,m,x-1));
  }
  return 0;
}

  

 

相關文章