qscoj 66 ||2017 UESTC Training for Data Structures D(离线+树状数组)

1.qscoj 66

离线+树状数组。

询问，如果只有A数组的话，实际上就是权值线段树或者主席树的裸题了。

那么我们其实只要将询问按照A数组从小到大排序，然后依次删除对于>A不合法的，然后用个权值树状数组去查询，就可以了。

代码：

#include <bits/stdc++.h>
using  namespace  std;

#define ff first
#define ss second
#define pb push_back
#define ll long long
#define mod 1000000007
#define ull unsigned long long
#define min3(a, b, c) min(a, min(b, c))
#define max3(a, b, c) max(a, max(b, c))
#define mst(ss,b) memset(ss,b,sizeof(ss));
#define dbg(x) cout << #x << "= " << x << endl;
typedef pair <int, int> pii;
const int inf = 0x3f3f3f3f;
const ll INF = (1LL<<63)-1;
const int N = 1e5+5;

int n, Q;
pair<int, int>a
;
struct node{
int x, y, id;
bool operator < (const node& rhs)const {
return x < rhs.x;
}
}q
;
int ans
;
struct BIT{
int n;
ll c
;
void init(int _n){
n = _n;
for(int i=1; i<=n; i++)c[i] = 0;
}
void update(int p, int x){
for(int i=p; i<=n; i+=i&-i)c[i] += x;
}
int query(int p){
int ret = 0;
for(int i=p; i>0; i-=i&-i)ret += c[i];
return ret;
}
}bit;

int  main(){
while(~scanf("%d%d", &n, &Q)){
int tol = 0;
mst(ans, 0);
bit.init(n);
for(int i=1; i<=n; i++)scanf("%d", &a[i].ff);
for(int i=1; i<=n; i++){
scanf("%d", &a[i].ss);
tol++;
bit.update(a[i].ss, 1);
}
for(int i=1; i<=Q; i++){
scanf("%d%d", &q[i].x, &q[i].y);
q[i].id = i;
}
sort(a+1, a+1+n);
sort(q+1, q+1+Q);
int now = 1;
for(int i=1; i<=Q; i++){
while(now <= n && a[now].ff < q[i].x){
bit.update(a[now].ss, -1);
tol--;
now++;
}
ans[q[i].id] = tol-bit.query(q[i].y-1);
}
for(int i=1; i<=Q; i++)printf("%d\n", ans[i]);
}
return 0;
}

2017 UESTC Training for Data Structures D

做法同上题

#include <set>
#include <map>
#include <stack>
#include <queue>
#include <string>
#include <vector>
#include <math.h>
#include <time.h>
#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <string.h>
#include <algorithm>
using  namespace  std;

#define ff first
#define ss second
#define pb push_back
#define ll long long
#define mod 1000000007
#define ull unsigned long long
#define min3(a, b, c) min(a, min(b, c))
#define max3(a, b, c) max(a, max(b, c))
#define mst(ss,b) memset(ss,b,sizeof(ss));
#define dbg(x) cout << #x << "= " << x << endl;
typedef pair <int, int> pii;
const int inf = 0x3f3f3f3f;
const ll INF = (1LL<<63)-1;
const int N = 1e5+5;

pii a
;
struct node{
int x, y, id;
bool operator < (const node& rhs)const{
return x < rhs.x;
}
}q
;

struct BIT{
int n;
ll c
;
void init(int _n){
n = _n;
for(int i=1; i<=n; i++)c[i] = 0;
}
void update(int p, int x){
for(int i=p; i<=n; i+=i&-i)c[i] += x;
}
int query(int p){
int ret = 0;
for(int i=p; i>0; i-=i&-i)ret += c[i];
return ret;
}
}bit;
int cnt
, ans
;

int  main(){
int n;
scanf("%d", &n);
bit.init(n);
mst(cnt, 0);
mst(ans, 0);
for(int i=1; i<=n; i++){
scanf("%d%d", &a[i].ff, &a[i].ss);
bit.update(a[i].ss, 1);
q[i].x = a[i].ff, q[i].y = a[i].ss;
q[i].id = i;
}
sort(a+1, a+1+n);
sort(q+1, q+1+n);
int now = n;
for(int i=n; i>=1; i--){
while(now >= 1 && a[now].ff > q[i].x){
bit.update(a[now].ss, -1);
now--;
}
ans[q[i].id] = bit.query(q[i].y)-1;
}
//    for(int i=1; i<=n; i++)dbg(ans[i]);
for(int i=1; i<=n; i++)cnt[ans[i]]++;
for(int i=1; i<=n; i++)printf("%d\n", cnt[i-1]);
return 0;
}