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[N]) 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;}