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QOJ
| ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
|---|---|---|---|---|---|---|---|---|---|
| #662022 | #7007. Rikka with Data Structures | pengpeng_fudan | WA | 3376ms | 23784kb | C++23 | 5.9kb | 2024-10-20 20:08:33 | 2024-10-20 20:08:44 |
Judging History
answer
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define int long long
int a[100010];
int n,m;
struct seg{//楼房重建(pushup(log)),解决计算前缀下最小值/最大值位置数量问题
#define mx(x) tr[x].mx
#define num(x) tr[x].num
#define tag1(x) tr[x].tag1
#define tag2(x) tr[x].tag2
struct node{
int mx;int num;//num(x)表示只考虑这一段(l,r)时的答案
int tag1;int tag2;
}tr[400010];
void build(int p,int l,int r,vector<int>& ve){
mx(p)=num(p)=tag1(p)=tag2(p)=0;
if(l==r){
mx(p)=ve[l];num(p)=1;
return ;
}
int mid=(l+r)>>1;
build(p<<1,l,mid,ve);build(p<<1|1,mid+1,r,ve);
pushup1(p);
num(p)=get(p,l,r);
}
void pushup1(int p){
mx(p)=max(mx(p<<1),mx(p<<1|1));
}
int pushup2(int p,int l,int r,int len){
if(l==r) return mx(p)>=len;
spread(p,l,r);
if(mx(p)<len) return 0;
int mid=(l+r)>>1;
if(mx(p<<1|1)>=len) return num(p)-num(p<<1|1)+pushup2(p<<1|1,mid+1,r,len);
else return pushup2(p<<1,l,mid,len);
}
void spread(int p,int l,int r){
tag1(p<<1)+=tag1(p),tag1(p<<1|1)+=tag1(p);
mx(p<<1)+=tag1(p);mx(p<<1|1)+=tag1(p);
tag1(p)=0;
if(tag2(p)){
int mid=(l+r)>>1;
mx(p<<1)=tag2(p),mx(p<<1|1)=tag2(p);num(p<<1)=mid-l+1,num(p<<1|1)=r-mid;
tag2(p<<1)=tag2(p),tag2(p<<1|1)=tag2(p);
tag2(p)=0;
}
}
int get(int p,int l,int r){
int mid=(l+r)>>1;spread(p,l,r);
return num(p<<1|1)+((mx(p<<1)<mx(p<<1|1))?0:pushup2(p<<1,l,mid,mx(p<<1|1)));
}
void modify(int p,int l,int r,int L,int R,int ad,int t){//0覆盖，1加法
if(l!=r) spread(p,l,r);
if(L<=l&&r<=R){
if(t==0) {tag2(p)=ad;mx(p)=ad,num(p)=r-l+1;}
else {
tag1(p)+=ad;
mx(p)+=ad;
}
return ;
}
int mid=(l+r)>>1;
if(mid>=L) modify(p<<1,l,mid,L,R,ad,t);
if(mid<R) modify(p<<1|1,mid+1,r,L,R,ad,t);
pushup1(p);
num(p)=get(p,l,r);
}
void modify(int l,int r,int num,int t){
modify(1,1,n,l,r,num,t);
}
int queryL(int p,int l,int r,int L,int R,int nz){
if(l==r){
return mx(p)>nz?l:-1;
}
spread(p,l,r);
int mid=(l+r)>>1;
if(mid<R&&mx(p<<1|1)>=nz){
int t=queryL(p<<1|1,mid+1,r,L,R,nz);
if(t!=-1) return t;
}
return queryL(p<<1,l,mid,L,R,nz);
}
int queryL(int pz,int num){//pz左边第一个比它大的位置
if(pz==1) return -1;
return queryL(1,1,n,1,pz-1,num);
}
int ask(int p,int l,int r,int L,int R){
if(L<=l&&r<=R) return mx(p);
spread(p,l,r);
int mid=(l+r)>>1;
if(mid>=L&&mid<R) return max(ask(p<<1,l,mid,L,R),ask(p<<1|1,mid+1,r,L,R));
if(mid>=L) return ask(p<<1,l,mid,L,R);
if(mid<R) return ask(p<<1|1,mid+1,r,L,R);
return -1;
}
int ask(int l,int r){
return ask(1,1,n,l,r);
}
int query_stack_num(int p,int l,int r,int L,int R,int nz){
if(L<=l&&r<=R) {
return pushup2(p,l,r,nz);
}
int mid=(l+r)>>1;
if(mid>=L&&mid<R) {
int nxtl=max(nz,mx(p>>1|1));
int ans=query_stack_num(p<<1,l,mid,L,R,nxtl)+query_stack_num(p<<1|1,mid+1,r,L,R,nz);
return ans;
// return query_stack_num(p<<1,l,mid,L,R,nxtl)+query_stack_num(p<<1|1,mid+1,r,L,R,nz);
}
if(mid>=L) return query_stack_num(p<<1,l,mid,L,R,nz);
if(mid<R) return query_stack_num(p<<1|1,mid+1,r,L,R,nz);
return -1;
}
int query_stack_num(int l,int r,int nz){
return query_stack_num(1,1,n,l,r,nz);
}
};
seg sg;
void solve(){
cin>>n>>m;
vector<int> a(n+1);
for(int i=1;i<=n;i++){
cin>>a[i];
}
struct node{
int op,l,r,x;
};
vector<node> ak(m+1);
vector<int> ans(m+1,0);
for(int i=1;i<=m;i++){
cin>>ak[i].op>>ak[i].l>>ak[i].r>>ak[i].x;
if(n==321) cout<<ak[i].op<<' '<<ak[i].l<<' '<<ak[i].r<<' '<<ak[i].x<<'\n';
}
auto ope=[&](vector<int>& ve,vector<node>& qry)->void {
sg.build(1,1,n,ve);
// cerr<<'\n';
for(int i=1;i<=m;i++){
auto [op,l,r,x]=qry[i];
if(op==1){
sg.modify(l,r,x,1);
}
else if(op==2){
sg.modify(l,r,x,0);
}
else if(l!=-1&&l<=r){
int w=sg.ask(x,x);
int pz=sg.queryL(x,w);
ans[i]+=max(r-max(l,pz)+1,0ll);
if(pz>l) ans[i]+=sg.query_stack_num(l,min(r,pz-1),sg.ask(pz,pz));
}
}
};
vector<node> qry1(m+1),qry2(m+1);
for(int i=1;i<=m;i++){
if(ak[i].op!=3) qry1[i]=qry2[i]=ak[i];
else{
auto [op,l,r,x]=ak[i];
if(x<=r&&x>=l) {
ans[i]++;
qry1[i]={3,l,x-1,x};
qry2[i]={3,x+1,r,x};
}
else if(x<l){
qry1[i]={3,-1,-1,x};
qry2[i]={3,l,r,x};
}
else{
qry1[i]={3,l,r,x};
qry2[i]={3,n+2,n+2,x};
}
}
}
for(auto& [op,l,r,x]:qry2){
l=n-l+1,r=n-r+1;swap(l,r);
if(op==3) x=n-x+1;
}
ope(a,qry1);
reverse(a.begin()+1,a.end());
ope(a,qry2);
for(int i=1;i<=m;i++){
if(ak[i].op==3) cout<<ans[i]<<'\n';
}
}
signed main() {
ios::sync_with_stdio(false), cin.tie(nullptr);
int _ = 1;
cin >> _;
while (_--) solve();
return 0;
}
Details
Tip: Click on the bar to expand more detailed information
Test #1:
score: 100
Accepted
time: 0ms
memory: 3584kb
input:
1 10 10 1 3 2 5 2 3 1 6 4 5 3 5 7 8 3 5 7 4 1 1 5 2 3 1 10 4 3 1 10 8 2 8 8 8 3 1 10 8 3 1 10 4 2 4 8 1 3 1 2 10
output:
3 3 10 7 10 8 2
result:
ok 7 lines
Test #2:
score: -100
Wrong Answer
time: 3376ms
memory: 23784kb
input:
200 321 43 168330405 102091681 86278243 227886812 609333939 211045240 332465535 212315420 510322126 700237719 102348318 320419595 409640374 582249257 245617532 643949598 748863235 405762764 358055464 585833725 429993246 60296212 632603910 229141445 696836739 297883078 545245133 44079558 92873286 347...
output:
2 81 145 286160135 3 19 52 183 1 15 59 3209535 3 93 220 105 2 113 186 186254041 3 100 169 172 1 54 94 414671130 2 58 227 282009609 3 37 242 285 1 105 231 283914614 1 59 260 327800140 2 20 67 384075682 3 92 233 262 2 121 254 646763759 1 85 143 589993469 2 171 223 437544961 2 210 228 446164910 2 88 23...
result:
wrong answer 1st lines differ - expected: '1', found: '2 81 145 286160135'