HDU 4217 Data Structure?（线段树的查找和更新）

Problem Description

Data structure is one of the basic skills for Computer Science students, which is a particular way of storing and organizing data in a computer so that it can be used efficiently. Today let me introduce a data-structure-like problem for you.

Original, there are N numbers, namely 1, 2, 3…N. Each round, iSea find out the Ki-th smallest number and take it away, your task is reporting him the total sum of the numbers he has taken away.

Input

The first line contains a single integer T, indicating the number of test cases.

Each test case includes two integers N, K, K indicates the round numbers. Then a line with K numbers following, indicating in i (1-based) round, iSea take away the Ki-th smallest away.

Technical Specification

1. 1 <= T <= 128

2. 1 <= K <= N <= 262 144

3. 1 <= Ki <= N - i + 1

Output

For each test case, output the case number first, then the sum.

Sample Input

2

3 2

1 1

10 3

3 9 1

Sample Output

Case 1: 3

Case 2: 14

本题是简单的线段树的更新和查找题目；

具体的步骤在代码中打上了注释。

下面是AC代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
using namespace std;

struct node
{
ll right,left,val;
}c[600005];

void build_tree(ll root,ll l,ll r)//建树
{
c[root].left=l;
c[root].right=r;
c[root].val=r-l+1;//表示这个节点有多少个数
if(c[root].left==c[root].right)
{
return ;
}
ll mid=(l+r)/2;
build_tree(root*2,l,mid);
build_tree(root*2+1,mid+1,r);
}

int find_tree(ll root,ll k)//查找第k小的数
{
c[root].val--;//每查找一次，这个节点上的数就少一个
if(c[root].left==c[root].right)
{
return c[root].left;
}
if(c[root*2].val>=k)//如果左子节点上的数的个数大于k,那么只需要搜索左子节点
find_tree(root*2,k);
else//否则的话搜索右子节点
{
find_tree(root*2+1,k-c[root*2].val);//至于这里为什么用k来减，原因很明确，就是左子节点有c[root*2].val个值
}
}

int main()
{
int t;
ll n,m;
int kase=0;
scanf("%d",&t);
while(t--)
{
kase++;
scanf("%lld%lld",&n,&m);
build_tree(1,1,n);
ll num,re=0;
for(int i=0;i<m;i++)
{
scanf("%lld",&num);
re+=find_tree(1,num);
}
printf("Case %d: %lld\n",kase,re);
}
return 0;
}