洛谷3348
可持久化线段树模板
1.结构体的打法
#include<cstdio> #include<cctype> #include<algorithm> using namespace std; const int maxn=200010; struct data{int l,r,sz;}tr[maxn<<5]; int n,m,cnt,key,a[maxn],rk[maxn],id[maxn],rt[maxn]; inline int read(int &x){ char ch=getchar();x=0; while(!isdigit(ch))ch=getchar(); while(isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();} } bool cmp(const int &x,const int &y){ return a[x]<a[y]; } void build(int l,int r,int &pos){ tr[++cnt]=tr[pos];pos=cnt;tr[cnt].sz++; if(l==r)return; int mid=(l+r)>>1; if(key<=mid)build(l,mid,tr[cnt].l);else build(mid+1,r,tr[cnt].r); } int query(int l,int r,int x,int y,int k){ if(l==r)return l; int mid=(l+r)>>1,sz=tr[tr[y].l].sz-tr[tr[x].l].sz; if(k<=sz)return query(l,mid,tr[x].l,tr[y].l,k); else return query(mid+1,r,tr[x].r,tr[y].r,k-sz); } int main(){ read(n);read(m); for(int i=1;i<=n;i++){read(a[i]);id[i]=i;} sort(id+1,id+n+1,cmp); for(int i=1;i<=n;i++)rk[id[i]]=i; for(int i=1;i<=n;i++){ rt[i]=rt[i-1]; key=rk[i];build(1,n,rt[i]); } while(m--){ int l,r,k;read(l);read(r);read(k); printf("%d\n",a[id[query(1,n,rt[l-1],rt[r],k)]]); } }
2.数组打法
#include<cstdio> #include<cctype> #include<algorithm> #define maxn 200005 using namespace std; int n,m,cnt,rt[maxn],rk[maxn]; int l[maxn<<5],r[maxn<<5],siz[maxn<<5],a[maxn],id[maxn]; inline void read(int &x){ char ch=getchar();x=0; while(!isdigit(ch))ch=getchar(); while(isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();} } inline void buildtr(int &root,int L,int R,int now){ l[++cnt]=l[root];r[cnt]=r[root];siz[cnt]=siz[root]+1; root=cnt; if(L==R)return; int mid=(L+R)>>1; if(now<=mid)buildtr(l[cnt],L,mid,now);else buildtr(r[cnt],mid+1,R,now); } inline int query(int L,int R,int x,int y,int k){ if(L==R)return L; int mid=(L+R)>>1,sz=siz[l[y]]-siz[l[x]]; if(sz>=k)return query(L,mid,l[x],l[y],k);else return query(mid+1,R,r[x],r[y],k-sz); } inline bool cmp(int x,int y){ return a[x]<a[y]; } int main(){ read(n);read(m); for(int i=1;i<=n;i++){read(a[i]);id[i]=i;} sort(id+1,id+n+1,cmp); for(int i=1;i<=n;i++)rk[id[i]]=i; for(int i=1;i<=n;i++){ rt[i]=rt[i-1];buildtr(rt[i],1,n,rk[i]); } while(m--){ int L,R,k;read(L);read(R);read(k);printf("%d\n",a[id[query(1,n,rt[L-1],rt[R],k)]]); } }