hdu2665

题解:

裸的主席树,记录最小值

代码:

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
using namespace std;
const int N=100005;
int a[N],cnt,Case,rt[N],b[N],now,n,m,x,y,k;
struct Tree
{
    int ls,rs,num;
}T[N*20];
int ef(int x)
{
    int l=1,r=cnt;
    while (l<r)
     {
         int mid=(l+r)/2;
         if (b[mid]<x)l=mid+1;
         else r=mid;
     }
    return l; 
}
int build(int l,int r)
{
    now++;
    if (l==r)return now;
    int k=now,mid=(l+r)/2;
    if (l<=mid)T[k].ls=build(l,mid);
    if (mid<r)T[k].rs=build(mid+1,r);
    return k; 
}
int insert(int x,int l,int r,int s)
{
    now++;
    if (l==r)
     {
         T[now].num=T[x].num+1;
         return now;
     }
    int k=now,mid=(l+r)/2;
    if (s<=mid)
     {
         T[k].ls=insert(T[x].ls,l,mid,s);
         T[k].rs=T[x].rs;
     }
    else
     {
         T[k].ls=T[x].ls;
         T[k].rs=insert(T[x].rs,mid+1,r,s);
     } 
    T[k].num=T[T[k].ls].num+T[T[k].rs].num;
    return k; 
}
int find(int x,int y,int z,int l,int r)
{
    if (l==r)return l;
    if (T[T[x].ls].num-T[T[y].ls].num>=z)
     return find(T[x].ls,T[y].ls,z,l,(l+r)/2);
    return find(T[x].rs,T[y].rs,z-
    T[T[x].ls].num+T[T[y].ls].num,(l+r)/2+1,r);
}
int main()
{
    scanf("%d",&Case); 
    while (Case--)
     {
         memset(T,0,sizeof T);
         now=0;
         scanf("%d%d",&n,&m);
         for (int i=1;i<=n;i++)scanf("%d",&a[i]);
         for (int i=1;i<=n;i++)b[i]=a[i];
         sort(b+1,b+n+1);
         cnt=1;
         for (int i=2;i<=n;i++)
          if (b[i]!=b[i-1])b[++cnt]=b[i];
         for (int i=1;i<=n;i++)a[i]=ef(a[i]);
        rt[0]=build(1,n);
        for (int i=1;i<=n;i++)rt[i]=insert(rt[i-1],1,n,a[i]);
        while (m--)
         {
             scanf("%d%d%d",&x,&y,&k);
             printf("%d
",b[find(rt[y],rt[x-1],k,1,n)]);
         } 
     }
}
原文地址:https://www.cnblogs.com/xuanyiming/p/7900412.html