Boring Counting
Time Limit: 3000MS Memory Limit: 65536KB
Problem Description
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).
Input
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)
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)
Output
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.
Example Input
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
Example Output
Case #1: 13 7 3 6 9
题目链接：SDUT 2610
裸的主席树区间查询，但是要注意一下就是离散化之后若输入的值域中有小于最小值的，即计算原位置后会出现0，因此建树若范围是1~pos.size()的话就会超时，建议改为0~pos.size()；
若有大于最大值的，则离散化之后会自动替换被为最大值，然而此题这样做当然是错的，因此a不用找接近值代替而要把上限b要找接近值代替，这样一来若b大于最大值，则代替没有关系，若b等于a且均大于最大值，则会出现b比a小的情况，特判一下输出0，然后若b比最小值都要小，则直接让B等于0，也特判为输出0。
调试了一会儿终于过了，= =划分树什么的这题肯定木有这个主席树这么好用啦……再说数据结构还没学到这玩意儿……也就会个主席树了……。
最后把这组数据调通就过了……
1
13 8
1 2 3 4 5 6 7 8 9 10 11 12 13
1 13 14 14
答案应该是0，如果答案是1的话估计就是离散化位置计算出了点问题
代码：
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <sstream>
#include <cstring>
#include <bitset>
#include <string>
#include <deque>
#include <stack>
#include <cmath>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int N=50010;
struct seg
{
int lson,rson;
int cnt;
};
seg T[N*20];
int root[N],tot;
pii arr[N];
vector<int>pos;
void update(int &cur,int ori,int l,int r,int pos)
{
cur=++tot;
T[cur]=T[ori];
++T[cur].cnt;
if(l==r)
return ;
int mid=MID(l,r);
if(pos<=mid)
update(T[cur].lson,T[ori].lson,l,mid,pos);
else
update(T[cur].rson,T[ori].rson,mid+1,r,pos);
}
int query(int S,int E,int l,int r,int L,int R)
{
if(L<=l&&r<=R)
return T[E].cnt-T[S].cnt;
else
{
int mid=MID(l,r);
if(R<=mid)
return query(T[S].lson,T[E].lson,l,mid,L,R);
else if(L>mid)
return query(T[S].rson,T[E].rson,mid+1,r,L,R);
else
return query(T[S].lson,T[E].lson,l,mid,L,mid)+query(T[S].rson,T[E].rson,mid+1,r,mid+1,R);
}
}
void init()
{
CLR(root,0);
tot=0;
T[0].cnt=T[0].lson=T[0].rson=0;
pos.clear();
}
int main(void)
{
int tcase,n,m,l,r,a,b,i;
scanf("%d",&tcase);
for (int q=1; q<=tcase; ++q)
{
init();
scanf("%d%d",&n,&m);
for (i=1; i<=n; ++i)
{
scanf("%d",&arr[i].first);
pos.push_back(arr[i].first);
}
sort(pos.begin(),pos.end());
pos.erase(unique(pos.begin(),pos.end()),pos.end());
int SZ=pos.size();
for (i=1; i<=n; ++i)
{
arr[i].second=lower_bound(pos.begin(),pos.end(),arr[i].first)-pos.begin()+1;
update(root[i],root[i-1],0,SZ,arr[i].second);
}
printf("Case #%d:
",q);
for (i=0; i<m; ++i)
{
scanf("%d%d%d%d",&l,&r,&a,&b);
int A,B;
A=lower_bound(pos.begin(),pos.end(),a)-pos.begin()+1;
B=lower_bound(pos.begin(),pos.end(),b)-pos.begin()+1;
if(pos[B-1]!=b)//上限B的位置要特别计算一下
--B;
if(!B||A>B)//特判
puts("0");
else
printf("%d
",query(root[l-1],root[r],0,SZ,A,B));
}
}
return 0;
}