Boring Counting——【SDUT2610】主席树

Boring Counting

Time Limit: 3000ms Memory limit: 65536K

题目描述

In this problem you are given a number sequence P consisting of N integer and Pi is the ith element in the sequence. Now you task is to answer a list of queries, for each query, please tell us among [L, R], how many Pi is not less than A and not greater than B( L<= i <= R). In other words, your task is to count the number of Pi (L <= i <= R, A <= Pi <= B).

输入

In the first line there is an integer T (1 < T <= 50), indicates the number of test cases.

For each case, the first line contains two numbers N and M (1 <= N, M <= 50000), the size of sequence P, the number of queries. The second line contains N numbers Pi(1 <= Pi <= 10^9), the number sequence P. Then there are M lines, each line contains four number L, R, A, B(1 <= L, R <= n, 1 <= A, B <= 10^9)

输出

For each case, at first output a line ‘Case #c:’, c is the case number start from 1. Then for each query output a line contains the answer.

示例输入

1

13 5

6 9 5 2 3 6 8 7 3 2 5 1 4

1 13 1 10

1 13 3 6

3 6 3 6

2 8 2 8

1 9 1 9

示例输出

Case #1:

13

7

3

6

9

来源

2013年山东省第四届ACM大学生程序设计竞赛

第一次敲主席树，之前感觉主席树好高大上，现在自己认真的去实现的时候会发现主席树很有趣。主席树的介绍在网上已经由很多了，不在介绍，直接上代码，或许代码中的注释会更加的清楚。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <string>
#include <queue>
#include <stack>
#include <set>
#include <algorithm>
#include <iostream>

using namespace std;

typedef long long LL;

const int Max = 1e5;

typedef struct node
{
int l,r;

int data;
}Tree;

Tree Tr[Max*10];

int num[Max],srt[Max];

int root[Max];

int top;

int Creat()
{
Tr[top].l = Tr[top].r = -1;

Tr[top].data = 0 ;

return top++ ;
}

int BS(int *a, int L, int R, int goal) //二分
{
int ans = -1;

while(L<=R)
{
int mid = (L+R) >> 1;

if(a[mid] == goal) return mid;

if(a[mid]<goal) L=mid+1,ans = mid;

else R =  mid-1;
}
return ans;
}

void Init(int &fa,int l,int r)//初始化，建立0状态时的线段树。
{
fa = Creat();

if(l == r) return ;

int mid = (l+r) >> 1;

Init(Tr[fa].l,l,mid);

Init(Tr[fa].r,mid+1,r);
}

void Update(int pre,int now,int l,int r,int goal)//更新其他的状态
{
if(l == r)
{
Tr[now].data = Tr[pre].data+1;

return ;
}

int mid = (l + r) >> 1;

if(goal <= mid)
{
Tr[now].r = Tr[pre].r;

Tr[now].l = Creat();

Update(Tr[pre].l,Tr[now].l,l,mid,goal);
}
else
{
Tr[now].l = Tr[pre].l;

Tr[now].r = Creat();

Update(Tr[pre].r,Tr[now].r,mid+1,r,goal);
}

Tr[now].data = Tr[Tr[now].l].data+Tr[Tr[now].r].data;
}

int Query(int now,int l,int r ,int goal)//查询在[0,now]的区间中的<= goal 的数的数量
{
if(goal==-1)
{
return 0;
}

if(r==goal)
{
return Tr[now].data;
}

int mid = (l + r) >> 1;

if(mid < goal) return Tr[Tr[now].l].data+Query(Tr[now].r,mid+1,r,goal);

else return Query(Tr[now].l, l, mid, goal);

}

int main()
{
int T,z = 1;

int n,m;

scanf("%d",&T);

while(T--)
{
scanf("%d %d",&n,&m);

for(int i = 1 ; i <= n;i++)
{
scanf("%d",&num[i]);

srt[i] = num[i];
}

sort(srt+1, srt + n + 1);

int N = 1;

for(int i = 2;i <= n; i++)//离散化
{
if(srt[i] != srt
) srt[++N] = srt[i];
}

for(int i = 1;i<=n;i++)
{
num[i] = BS(srt,1,N,num[i]);

}

root[0] = -1;

top = 0;

Init(root[0],1,N);//初始化主席树

for(int i = 1;i<=n;i++)//更新状态
{
root[i] = Creat();

Update(root[i-1],root[i],1,N,num[i]);
}

int l,r,Mi,Ma;

printf("Case #%d:\n",z++);

while(m--)
{
scanf("%d %d %d %d",&l,&r,&Mi,&Ma);

if(l>r||Mi>Ma)
{
printf("0\n");

continue;
}

Mi = BS(srt,1,N,Mi-1);//找到所要查询的区间

Ma = BS(srt,1,N,Ma);

printf("%d\n",Query(root[r],1,N,Ma)-Query(root[r],1,N,Mi)-Query(root[l-1],1,N,Ma)+Query(root[l-1],1,N,Mi));
}
}
return 0;
}