TJU 3758. Jewel 【划分树，树状数组】

http://acm.tju.edu.cn/toj/showp3758.html
TJU 3758.   Jewel
模拟题，开始数组为空，有以下操作：
Insert x
    Put a bead with size x to the right of the chain (0 < x < 231, and x is different from all the sizes of beads currently in the chain)

Query_1 s t k
    Query the k-th smallest bead between position s and t, inclusive. You can assume 1 ≤ s ≤ t ≤ L, (L is the length of the current chain), and 1 ≤ k ≤ min (100, t - s + 1)

Query_2 x
    Query the rank of the bead with size x, if we sort all the current beads by ascent order of sizes. The result should between 1 and L (L is the length of the current chain)

Query_3 k
    Query the size of the k-th smallest bead currently (1 ≤ k ≤ L, L is the length of the current chain)

要求统计每种问题的答案

分析：对于问题1和3可用划分树解决，对于问题2用树状数组解决，注意用树状数组前需要对数据离散化

#include<stdio.h>
#include<algorithm>
#include<iostream>
using namespace std;
typedef long long LL;
struct Q//问题
{
  int kind;//问题类型
  int x,k;
  int s,t;
  int ans;
}q[35000*4+10];
const int MAXN = 100000+10;
int tree[22][MAXN],toleft[22][MAXN];
int sorted[MAXN];//已经排好序的数据
int c[MAXN];
inline int low(int x)
{
 return (x&(-x));
}
void add(int x, int n)
{
 while( x <= n)
 {
  c[x] ++;
  x += low(x);
 }
}
LL Sum(int x)
{
 LL ans = 0;
 while(x > 0)
 {
  ans += c[x];
  x -= low(x);
 }
 return ans;
}

void build_tree(int left,int right,int deep)//建树
{
 if(left==right)return;
 int mid = (left+right)>>1;
 int i;
 int same = mid-left+1;//位于左子树的数据
 for(i=left;i<=right;i++)  //计算放于左子树中与中位数相等的数字个数
 {
  if(tree[deep][i]<sorted[mid])
   same--;
 }

 int ls = left;
 int rs = mid+1;
 for(i=left;i<=right;i++)
 {
  int flag = 0;
  if((tree[deep][i]<sorted[mid])||(tree[deep][i]==sorted[mid]&&same>0))
  {
   flag = 1;
   tree[deep+1][ls++] = tree[deep][i];
   if(tree[deep][i]==sorted[mid])same--;
  }
  else
  {
   tree[deep+1][rs++] = tree[deep][i];
  }
      
  toleft[deep][i]=toleft[deep][i-1]+flag;
 }

 build_tree(left,mid,deep+1);
 build_tree(mid+1,right,deep+1);
}

int query(int left,int right,int k,int L,int R,int deep)
{
 if(left==right)return tree[deep][left];
    int mid = (L+R)>>1;
 int x = toleft[deep][left-1]-toleft[deep][L-1];//位于left左边的放于左子树中的数字个数
 int y = toleft[deep][right] - toleft[deep][L-1];//到right为止位于左子树的个数
 int ry = right-L-y;//到right右边为止位于右子树的数字个数
 int cnt = y-x;//[left,right]区间内放到左子树中的个数
 int rx = left-L-x;//left左边放在右子树中的数字个数
 if(cnt>=k)
  return query(L+x,L+y-1,k,L,mid,deep+1);
 else
  return query(mid+rx+1,mid+1+ry,k-cnt,mid+1,R,deep+1);

}

int find(int x,int left,int right)
{
 int ans = 0;
    while(left<=right)
 {
  int mid = (left+right)>>1;
  if(x==sorted[mid])return mid;
  if(x<sorted[mid])
   right=mid-1;
  else
   left=mid+1;
 }
 return ans;
}
int main()
{
 int op = 0;
 int cases = 0;
 while(scanf("%d",&op)!=EOF)
 {
  cases++;
  int n = 1;
  int qnum = 0;//问题个数
 
  while(op--)
  {
   char str[20];
   scanf("%s",str);
   if(str[0]=='I')
   {
     scanf("%d",&sorted[n]);
     tree[0][n]=sorted[n];
     n++;
   }
   else
   {
    int len = strlen(str);
    if(str[len-1]=='1')
    {
                    q[qnum].kind = 1;
     scanf("%d%d%d",&q[qnum].s,&q[qnum].t,&q[qnum].k);
     qnum++;
    }
    else
     if(str[len-1]=='2')
     {
      q[qnum].kind = 2;
      scanf("%d",&q[qnum].x);
      q[qnum].t = n-1;
         qnum++;
     }
     else
      if(str[len-1]=='3')
      {
       q[qnum].kind = 3;
       scanf("%d",&q[qnum].k);
       q[qnum].s=1;
       q[qnum].t = n-1;
       qnum++;
      }
   }
  }
  sort(sorted+1,sorted+n);
  memset(c,0,sizeof(c));
        int i,pre = 1;
  build_tree(1,n-1,0);
  LL sum1=0,sum2=0,sum3=0;
  for(i=0;i<qnum;i++)
  {
   if(q[i].kind==1)
   {
    //query(int left,int right,int k,int L,int R,int deep)
               sum1+=query(q[i].s,q[i].t,q[i].k,1,n-1,0);
   
   }
   else
    if(q[i].kind==3)
    {
     sum3+=query(q[i].s,q[i].t,q[i].k,1,n-1,0);
    }
   else
   {
    for(;pre<=q[i].t;pre++)
    {
     int index = find(tree[0][pre],1,n-1);
     add(index,n-1);
    }
    int index=find(q[i].x,1,n-1);
    //printf("%ld\n",Sum(index));
    sum2+=Sum(index);
   }
  }
      printf("Case %d:\n",cases);
   cout<<sum1<<endl;
   cout<<sum2<<endl;
   cout<<sum3<<endl;
    //  printf("%I64d\n%I64d\n%I64d\n",sum1,sum2,sum3);
 }
 return 0;
}