BZOJ 3196 Tyvj 1730 二逼平衡樹 ——樹狀數組套主席樹

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【題目分析】

    聽說是樹套樹。（霧）

    怒寫樹狀數組套主席樹，然後就Rank1了。23333

    單點修改，區間查詢+k大數查詢=樹狀數組套主席樹。

【代碼】

#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath> #include <set>#include <map>#include <string>#include <algorithm>#include <vector>#include <iostream>#include <queue>using namespace std; #define maxn 50005#define mlog 20 int read(){    int x=0,f=1; char ch=getchar();    while (ch<‘0‘||ch>‘9‘) {if (ch==‘-‘) f=-1; ch=getchar();}    while (ch>=‘0‘&&ch<=‘9‘) {x=x*10+ch-‘0‘; ch=getchar();}    return x*f;} int n,m;int rt[maxn<<1],a[maxn],b[maxn<<1],cnt=0,tot=0,sum,rnk,TMP;int ls[maxn<<6],rs[maxn<<6],siz[maxn<<6];int L[maxn],R[maxn],opt[maxn],x[maxn],y[maxn],z[maxn]; void ins(int o1,int & o2,int l,int r,int x,int f){//    printf("ins %d %d %d %d %d %d\n",o1,o2,l,r,x,f);    o2=++tot;    siz[o2]=siz[o1]+f;    if (l==r) return ;    int mid=(l+r)/2;    if (x<=mid) rs[o2]=rs[o1],ins(ls[o1],ls[o2],l,mid,x,f);    else ls[o2]=ls[o1],ins(rs[o1],rs[o2],mid+1,r,x,f);    return ;} void ready(int l,int r){//  cout<<"ready for "<<l<<" "<<r<<endl;     l--;    L[0]=R[0]=0;    for (int j=l;j;j-=j&(-j)) L[++L[0]]=rt[j];    for (int j=r;j;j-=j&(-j)) R[++R[0]]=rt[j];//  cout<<"L "; for (int j=1;j<=L[0];++j) cout<<L[j]<<" "; cout<<endl;//  cout<<"R "; for (int j=1;j<=R[0];++j) cout<<R[j]<<" "; cout<<endl;//  cout<<"L "; for (int j=1;j<=L[0];++j) cout<<siz[L[j]]<<" "; cout<<endl;//  cout<<"R "; for (int j=1;j<=R[0];++j) cout<<siz[R[j]]<<" "; cout<<endl;} void cholef(){//  cout<<"choose left"<<endl;    for (int i=1;i<=L[0];++i) L[i]=ls[L[i]];    for (int i=1;i<=R[0];++i) R[i]=ls[R[i]];} void chorig(){//  cout<<"choose right"<<endl;    for (int i=1;i<=L[0];++i) L[i]=rs[L[i]];    for (int i=1;i<=R[0];++i) R[i]=rs[R[i]];} int taksum(){    int ret=0;    for (int i=1;i<=L[0];++i) ret-=siz[ls[L[i]]];    for (int i=1;i<=R[0];++i) ret+=siz[ls[R[i]]];    return ret;} int qrnk(int l,int r,int x){//  printf("qrnk %d %d %d\n",l,r,x);    if (l==r) return 1;    int mid=(l+r)/2,tmp=taksum();//  cout<<"take sum "<<tmp<<endl;    if (x<=mid)  return cholef(),qrnk(l,mid,x);    else return chorig(),tmp+qrnk(mid+1,r,x);} int qnum(int l,int r,int x){    if (l==r) return l;    int mid=(l+r)/2,tmp=taksum();    if (x<=tmp) return cholef(),qnum(l,mid,x);    else return chorig(),qnum(mid+1,r,x-tmp);} int main(){    n=read();m=read();    for (int i=1;i<=n;++i) scanf("%d",&a[i]),b[++cnt]=a[i];    for (int i=1;i<=m;++i)    {        scanf("%d%d%d",&opt[i],&x[i],&y[i]);        if (opt[i]!=3) scanf("%d",&z[i]);        if (opt[i]==3) b[++cnt]=y[i];        if (opt[i]==4||opt[i]==5) b[++cnt]=z[i];    }    sort(b+1,b+cnt+1);    cnt=unique(b+1,b+cnt+1)-b-1;     for (int i=1;i<=n;++i) a[i]=lower_bound(b+1,b+cnt+1,a[i])-b;    for (int i=1;i<=m;++i)    {        if (opt[i]==1||opt[i]==4||opt[i]==5)z[i]=lower_bound(b+1,b+cnt+1,z[i])-b;        if (opt[i]==3) y[i]=lower_bound(b+1,b+cnt+1,y[i])-b;    }//  cout<<endl<<endl;//  for (int i=1;i<=cnt;++i) cout<<i<<" ";cout<<endl;//  for (int i=1;i<=cnt;++i) cout<<b[i]<<" "; cout<<endl;//  for (int i=1;i<=m;++i) cout<<opt[i]<<" "<<x[i]<<" "<<y[i]<<" "<<z[i]<<endl;//  cout<<endl<<endl;    for (int i=1;i<=n;++i)        for (int j=i;j<=n;j+=j&(-j))            ins(rt[j],rt[j],1,cnt,a[i],1);    for (int i=1;i<=m;++i)    {        switch(opt[i])        {            case 1:                ready(x[i],y[i]);                printf("%d\n",qrnk(1,cnt,z[i]));            break;            case 3:                for (int j=x[i];j<=n;j+=j&(-j))                {                    ins(rt[j],rt[j],1,cnt,a[x[i]],-1);                    ins(rt[j],rt[j],1,cnt,y[i],1);                }                a[x[i]]=y[i];            break;            case 2:                ready(x[i],y[i]);                printf("%d\n",b[qnum(1,cnt,z[i])]);            break;            case 4:                ready(x[i],y[i]);                rnk=qrnk(1,cnt,z[i]);//              printf("rnk is %d\n",rnk);                ready(x[i],y[i]);                printf("%d\n",b[qnum(1,cnt,rnk-1)]);            break;            case 5:                ready(x[i],y[i]);                TMP=0;                TMP-=qrnk(1,cnt,z[i]);                ready(x[i],y[i]);                TMP+=qrnk(1,cnt,z[i]+1);//              printf("it s have %d\n",TMP);                ready(x[i],y[i]);                rnk=qrnk(1,cnt,z[i]);                ready(x[i],y[i]);                printf("%d\n",b[qnum(1,cnt,rnk+TMP)]);                         }    }}

　　

BZOJ 3196 Tyvj 1730 二逼平衡樹 ——樹狀數組套主席樹