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算是一道模板题了，可惜弱的一B的我并不会划分树，花了点时间学了下，回头A了这道题
3s的限时跑了2.8s也是醉了。。。
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).
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, 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.
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
Case #1:
1
3
7
3
6
9
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 50005;
int n,m;
int tree[20][maxn];
int lnum[20][maxn];
int sorted[maxn];
void debug(int v){
for(int i = 1; i <= n; i++) printf("%d ",sorted[i]);
puts("");
for(int i = 0; i < v; i++){
for(int j = 1; j <= n; j++)
printf("%d ",tree[i][j]);
puts("");
}
}
void build(int l,int r,int d){
if(l == r) return;
int mid = (l + r) >> 1;
int same = mid - l + 1;
for(int i = l; i <= r; i++)
if(tree[d][i] < sorted[mid])
same --;
int lpos = l,rpos = mid + 1;
lnum[d][0] = 0;
for(int i = l; i <= r; i++){
if(tree[d][i] < sorted[mid]){
lnum[d][i] = lnum[d][i - 1] + 1;
tree[d + 1][lpos++] = tree[d][i];
}
else if(tree[d][i] == sorted[mid] && same > 0){
same --;
lnum[d][i] = lnum[d][i - 1] + 1;
tree[d + 1][lpos++] = tree[d][i];
}
else{
lnum[d][i] = lnum[d][i - 1];
tree[d + 1][rpos++] = tree[d][i];
}
}
build(l,mid,d + 1);
build(mid + 1,r,d + 1);
}
int query(int L,int R,int l,int r,int d,int k){
//printf("%d %d %d %d\n",l,r,d,k);
if(l == r) return tree[d][l];
int mid = (L + R) >> 1;
int cnt = lnum[d][r] - lnum[d][l - 1];
if(cnt >= k){
int newl = L + lnum[d][l - 1] - lnum[d][L - 1];
int newr = newl + cnt - 1;
return query(L,mid,newl,newr,d + 1,k);
}
else{
int newr = r + lnum[d][R] - lnum[d][r];
int newl = newr - (r - l - cnt);
return query(mid + 1,R,newl,newr,d + 1,k - cnt);
}
}
void solve(){
int L,R,a,b;
scanf("%d%d%d%d",&L,&R,&a,&b);
int l = 1,r = (R - L + 2);
int v1 = 0,v2 = 0;
//找大于等于
while(l < r){
int mid = (l + r) >> 1;
int value = query(1,n,L,R,0,mid);
if(value < a)
l = mid + 1;
else if(value >= a)
r = mid;
}
v1 = l;
l = 1;
r = (R - L + 2);
//找大于
while(l < r){
int mid = (l + r) >> 1;
int value = query(1,n,L,R,0,mid);
if(value <= b){
l = mid + 1;
}
else if(value > b)
r = mid;
}
v2 = l;
printf("%d\n",v2 - v1);
}
int main(){
int T,Case = 1;
scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&m);
for(int i = 1; i <= n; i++){
scanf("%d",&tree[0][i]);
sorted[i] = tree[0][i];
}
sort(sorted + 1,sorted + n + 1);
build(1,n,0);
printf("Case #%d:\n",Case++);
for(int i = 0; i < m; i++)
solve();
}
return 0;
}
/*
5 5
1 1 1 1 1
1 5 2 3
2
*/
【山东省第四节ACM省赛】Boring Counting（二分+划分树）
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原文地址：http://blog.csdn.net/u013451221/article/details/45056557
