UPC 2224 Boring Counting (离线线段树,统计区间[l,r]之间大小在[A,B]中的数的个数)

时间:2023-03-09 13:03:53
UPC 2224 Boring Counting (离线线段树,统计区间[l,r]之间大小在[A,B]中的数的个数)

题目链接:http://acm.upc.edu.cn/problem.php?id=2224

题意:给出n个数pi,和m个查询,每个查询给出l,r,a,b,让你求在区间l~r之间的pi的个数(A<=pi<=B,l<=i<=r)。

参考链接:http://www.cnblogs.com/zj62/p/3558967.html

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
#define lson rt<<1,L,mid
#define rson rt<<1|1,mid+1,R
/*
http://acm.upc.edu.cn/problem.php?id=2224
*/
using namespace std;
const int maxn=+;
const int INF=0x3f3f3f3f;
int n,m;
int tree[maxn<<];
int ans[maxn][];
/*ans[i][0]记录第i个查询区间中比a小的数的个数,ans[i][1]记录第i个查询中比b小的数的个数,答案为ans[i][1]-ans[i][0]*/ struct Num{
int value;
int idx;
bool operator<(const Num tmp)const{
return value<tmp.value;
}
}num[maxn]; struct Query{
int l,r,a,b;
int idx;
}q[maxn]; bool cmp1(const Query tmp1,const Query tmp2){
return tmp1.a<tmp2.a;
}
bool cmp2(const Query tmp1,const Query tmp2){
return tmp1.b<tmp2.b;
} void build(int rt,int L,int R){
tree[rt]=;
if(L==R){
return;
}
int mid=(L+R)>>;
build(lson);
build(rson);
} void update(int rt,int L,int R,int x){
if(L==R){
tree[rt]++;
return;
}
int mid=(L+R)>>;
if(x<=mid)
update(lson,x);
else
update(rson,x);
tree[rt]=tree[rt<<]+tree[rt<<|];
} int query(int rt,int L,int R,int l,int r){
if(l<=L&&R<=r){
return tree[rt];
}
int mid=(L+R)>>;
int ret=;
if(l<=mid)
ret+=query(lson,l,r);
if(r>mid)
ret+=query(rson,l,r);
return ret;
} void solve(){
sort(num+,num+n+);
sort(q+,q+m+,cmp1);
build(,,n);
int d=;
for(int j=;j<=m;j++){
while(d<=n && num[d].value<q[j].a){
update(,,n,num[d].idx);
d++;
}
ans[q[j].idx][]=query(,,n,q[j].l,q[j].r);
}
sort(q+,q+m+,cmp2);
build(,,n);
d=;
for(int j=;j<=m;j++){
while(d<=n && num[d].value<=q[j].b){
update(,,n,num[d].idx);
d++;
}
ans[q[j].idx][]=query(,,n,q[j].l,q[j].r);
} }
int main()
{
int t,cases=;
scanf("%d",&t);
while(t--){
scanf("%d%d",&n,&m);
for(int i=;i<=n;i++){
scanf("%d",&num[i].value);
num[i].idx=i;
}
for(int i=;i<=m;i++){
scanf("%d%d%d%d",&q[i].l,&q[i].r,&q[i].a,&q[i].b);
q[i].idx=i;
}
solve();
printf("Case #%d:\n",++cases);
for(int i=;i<=m;i++){
printf("%d\n",ans[i][]-ans[i][]);
}
}
return ;
}