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线段树套pb_ds里的平衡树,在洛谷OJ上测试,后三个测试点TLE
#include<cstdio> #include<algorithm> #include<ext/pb_ds/assoc_container.hpp> #include<ext/pb_ds/tree_policy.hpp> #define N 50010 #define INF 2147483647 using namespace std; using namespace __gnu_pbds; typedef pair<int,int> Point; typedef tree<Point,null_type,less<Point>,rb_tree_tag,tree_order_statistics_node_update> rb_tree; typedef rb_tree::iterator ITER; rb_tree T[N<<2]; int a[N],tags[N],tag,n,m; void buildtree(int rt,int l,int r) { if(l==r) { T[rt].insert(Point(a[l],l)); return; } int m=(l+r>>1); buildtree(rt<<1,l,m); buildtree(rt<<1|1,m+1,r); T[rt]=T[rt<<1]; for(ITER it=T[rt<<1|1].begin();it!=T[rt<<1|1].end();++it) T[rt].insert(*it); } int Rank(int v,int TAG,int ql,int qr,int rt,int l,int r) { if(ql<=l&&r<=qr) return T[rt].order_of_key(Point(v,TAG)); int m=(l+r>>1),res=0; if(ql<=m) res+=Rank(v,TAG,ql,qr,rt<<1,l,m); if(m<qr) res+=Rank(v,TAG,ql,qr,rt<<1|1,m+1,r); return res; } int Pre(int v,int ql,int qr,int rt,int l,int r) { if(ql<=l&&r<=qr) { ITER it=T[rt].lower_bound(Point(v,0)); if(it==T[rt].begin()) return -INF; --it; return (*it).first; } int m=(l+r>>1),res=-INF; if(ql<=m) res=max(res,Pre(v,ql,qr,rt<<1,l,m)); if(m<qr) res=max(res,Pre(v,ql,qr,rt<<1|1,m+1,r)); return res; } int Nex(int v,int ql,int qr,int rt,int l,int r) { if(ql<=l&&r<=qr) { ITER it=T[rt].upper_bound(Point(v,INF)); if(it==T[rt].end()) return INF; return (*it).first; } int m=(l+r>>1),res=INF; if(ql<=m) res=min(res,Nex(v,ql,qr,rt<<1,l,m)); if(m<qr) res=min(res,Nex(v,ql,qr,rt<<1|1,m+1,r)); return res; } void Update(int p,int v,int rt,int l,int r) { T[rt].erase(Point(a[p],tags[p])); T[rt].insert(Point(v,tag)); if(l==r) return; int m=(l+r>>1); if(p<=m) Update(p,v,rt<<1,l,m); else Update(p,v,rt<<1|1,m+1,r); } int Kth(int K,int ql,int qr) { int l=0,r=100000000; while(l<r) { int m=(l+r>>1); if(Rank(m,INF,ql,qr,1,1,n)>=K) r=m; else l=m+1; } return l; } int main() { // freopen("bzoj3196.in","r",stdin); int op,x,y,z; scanf("%d%d",&n,&m); for(int i=1;i<=n;++i) { scanf("%d",&a[i]); tags[i]=i; } tag=n; buildtree(1,1,n); // for(int i=1;i<=n*4;++i) // printf("%d ",(*T[i].begin()).second); // puts(""); for(;m;--m) { scanf("%d%d%d",&op,&x,&y); if(op==1) { scanf("%d",&z); printf("%d\n",Rank(z,0,x,y,1,1,n)+1); } else if(op==2) { scanf("%d",&z); printf("%d\n",Kth(z,x,y)); } else if(op==3) { ++tag; Update(x,y,1,1,n); a[x]=y; tags[x]=tag; } else if(op==4) { scanf("%d",&z); printf("%d\n",Pre(z,x,y,1,1,n)); } else { scanf("%d",&z); printf("%d\n",Nex(z,x,y,1,1,n)); } } return 0; }
【线段树套平衡树】【pb_ds】bzoj3196 Tyvj 1730 二逼平衡树
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原文地址:http://www.cnblogs.com/autsky-jadek/p/5934325.html