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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#870043 | #7436. Optimal Ordered Problem Solver | Mirasycle | 0 | 2932ms | 125220kb | C++14 | 4.8kb | 2025-01-25 14:32:52 | 2025-01-25 14:33:54 |
Judging History
answer
#include<bits/stdc++.h>
#define pb emplace_back
#define fi first
#define se second
#define mp make_pair
using namespace std;
typedef long long ll;
const int maxn=1e6+10;
const int V=1e6;
void cmax(int &x,int y){ x=x>y?x:y; }
void cmin(int &x,int y){ x=x<y?x:y; }
struct BST{
struct node{
int r,s,ls,rs,tx,ty; pair<int,int> val;
#define val(x) tree[x].val
#define s(x) tree[x].s
#define ls(x) tree[x].ls
#define rs(x) tree[x].rs
#define r(x) tree[x].r
#define tx(x) tree[x].tx
#define ty(x) tree[x].ty
}tree[maxn]; int tot,root;
void init(){ tot=root=0; val(0)=mp(0,0); }
void pushup(int p){ s(p)=s(ls(p))+s(rs(p)); }
int New(pair<int,int> v){
int p=++tot; s(p)=1;
ls(p)=rs(p)=0; val(p)=v;
r(p)=rand(); return p;
}
void pushdown(int p){
if(tx(p)){
if(ls(p)) val(ls(p)).fi=tx(ls(p))=tx(p);
if(rs(p)) val(rs(p)).fi=tx(rs(p))=tx(p);
}
if(ty(p)){
if(ls(p)) val(ls(p)).se=ty(ls(p))=ty(p);
if(rs(p)) val(rs(p)).se=ty(rs(p))=ty(p);
}
tx(p)=ty(p)=0;
}
void split(int p,pair<int,int> k,int& x,int& y){
if(!p){ x=y=0; return ; }
pushdown(p);
if(val(p)<=k) x=p,split(rs(p),k,rs(p),y);
else y=p,split(ls(p),k,x,ls(p));
pushup(p);
}
void splitY(int p,int k,int &x,int& y){
if(!p){ x=y=0; return ; }
pushdown(p);
if(val(p).se<=k) x=p,splitY(rs(p),k,rs(p),y);
else y=p,splitY(ls(p),k,x,ls(p));
pushup(p);
}
int merge(int p,int q){
if(!p||!q) return p+q;
pushdown(p); pushdown(q);
if(r(p)<r(q)){
ls(q)=merge(p,ls(q));
pushup(q); return q;
}else{
rs(p)=merge(rs(p),q);
pushup(p); return p;
}
}
void ins(pair<int,int> v){
int x=0,y=0; split(root,v,x,y);
root=merge(merge(x,New(v)),y);
}
void upd(int o,int X,int Y){
int x=0,y=0,z=0;
split(root,mp(X,0),x,z);
splitY(x,-(Y+1),x,y);
if(o==1) val(y).se=Y,ty(y)=Y;
else val(y).fi=X,tx(y)=X;
root=merge(x,merge(y,z));
}
}t;
int n,m,tim[maxn],c[maxn],ans[maxn],b[maxn<<1];
struct BIT{
int c[maxn];
void init(int z){ for(int i=1;i<=V;i++) c[i]=z; }
int lowbit(int x){ return x&-x; }
void modify(int x,int v){ for(;x;x-=lowbit(x)) cmin(c[x],v); }
void modify2(int x,int v){ for(;x;x-=lowbit(x)) cmax(c[x],v); }
int sufmin(int x){ int res=m+1; for(;x<=V;x+=lowbit(x)) cmin(res,c[x]); return res; }
int sufmax(int x){ int res=0; for(;x<=V;x+=lowbit(x)) cmax(res,c[x]); return res; }
void add(int x,int v){ for(;x;x-=lowbit(x)) c[x]+=v; }
int query(int x){ int res=0; for(;x<=V;x+=lowbit(x)) res+=c[x]; return res; }
}t1,t2;
struct node{
int x,y;
}a[maxn]; vector<int> vec[maxn];
struct Q{
int o,x,y,X,Y;
}q[maxn];
int main(){
ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
cin>>n>>m; t.init();
for(int i=1;i<=n;i++) cin>>a[i].x>>a[i].y,c[a[i].x]++;
for(int i=1;i<=m;i++) cin>>q[i].o>>q[i].x>>q[i].y>>q[i].X>>q[i].Y,c[q[i].x]++;
//计算进入时间 tim
for(int i=1;i<=V;i++) c[i]+=c[i-1]; t1.init(m+1);
for(int i=1;i<=m;i++) b[c[q[i].x]--]=i;
for(int i=1;i<=n;i++) b[c[a[i].x]--]=i+m;
for(int i=n+m;i>=1;i--){
if(b[i]<=m) t1.modify(q[b[i]].y,b[i]);
else tim[b[i]-m]=t1.sufmin(a[b[i]-m].y);
}
//二维偏序 (x,t) (y,t) 注意 t,x,y 需要严格大于
memset(c,0,sizeof(c));
for(int i=1;i<=n;i++) c[tim[i]]++;
for(int i=1;i<=m;i++) c[i]+=c[i-1]+1; c[m+1]+=c[m];
for(int i=1;i<=m;i++) b[c[i]--]=i;
for(int i=1;i<=n;i++) b[c[tim[i]]--]=i+m;
t1.init(0); t2.init(0);
for(int i=n+m;i>=1;i--){
if(b[i]<=m) ans[b[i]]-=t1.query(q[b[i]].X+1),ans[b[i]]-=t2.query(q[b[i]].Y+1);
else t1.add(a[b[i]-m].x,1),t2.add(a[b[i]-m].y,1);
}
// for(int i=1;i<=n;i++) cout<<"A "<<ans[i]<<endl;
//二维偏序 (x,y) 还是需要严格大于
memset(c,0,sizeof(c));
for(int i=1;i<=n;i++) c[a[i].x]++;
for(int i=1;i<=m;i++) c[q[i].X]++;
for(int i=1;i<=V;i++) c[i]+=c[i-1]; t1.init(0);
for(int i=1;i<=m;i++) b[c[q[i].X]--]=i;
for(int i=1;i<=n;i++) b[c[a[i].x]--]=i+m;
for(int i=n+m;i>=1;i--){
if(b[i]<=m) ans[b[i]]+=n+t1.query(q[b[i]].Y+1);
else t1.add(a[b[i]-m].y,1);
}
// for(int i=1;i<=n;i++) cout<<"A "<<ans[i]<<endl;
//按照时间序维护轮廓线 查询
//同时需要维护当前轮廓线的范围,树状数组维护每个 x 对应的 y 即可 就是后缀max
for(int i=1;i<=n;i++) vec[tim[i]].pb(i); t1.init(0);
for(int i=1;i<=m;i++){
//加入平衡树中
t1.modify2(q[i].x,q[i].y);
t.upd(q[i].o,q[i].x,q[i].y);//修改
for(auto z:vec[i]){
if(q[i].o==1) t.ins(mp(a[z].x,q[i].y));
else t.ins(mp(q[i].x,a[z].x));
}
//查询
if(q[i].Y<t1.sufmax(q[i].X)){ ans[i]=0; continue; }
int x=0,y=0;
t.splitY(t.root,-(q[i].Y+1),x,y);
ans[i]-=t.s(x);
t.root=t.merge(x,y); x=y=0;
t.split(t.root,mp(q[i].X+1,0),x,y);
ans[i]-=t.s(y);
t.root=t.merge(x,y);
}
for(int i=1;i<=m;i++) cout<<ans[i]<<"\n";
return 0;
}
Details
Tip: Click on the bar to expand more detailed information
Subtask #1:
score: 0
Wrong Answer
Test #1:
score: 0
Wrong Answer
time: 7ms
memory: 76780kb
input:
995 996 950 481 376 18 141 776 711 966 130 727 468 529 47 39 857 563 832 821 776 472 154 914 825 279 332 415 470 361 968 440 45 560 299 755 703 785 744 387 547 382 3 549 344 573 778 424 784 765 280 115 251 434 695 98 463 506 379 38 610 486 305 623 703 244 856 365 117 360 772 847 331 723 663 991 900 ...
output:
423 473 758 313 694 232 333 784 821 154 247 244 504 54 200 370 872 782 734 174 660 467 76 754 77 5 144 974 526 415 439 694 507 577 258 120 243 9 8 319 313 498 218 828 433 623 981 700 120 55 70 499 375 283 387 128 317 139 220 410 22 547 20 385 430 168 38 0 178 625 677 561 488 672 577 64 144 537 235 6...
result:
wrong answer 26th numbers differ - expected: '3', found: '5'
Subtask #2:
score: 0
Wrong Answer
Test #9:
score: 0
Wrong Answer
time: 1461ms
memory: 118288kb
input:
999996 999996 921339 140126 955508 363759 958698 342406 280137 955674 907511 75721 189876 946586 152058 168837 541857 557702 84498 865987 185574 809224 704828 701026 815379 548000 989515 518951 335439 336866 745570 790616 766723 212893 926629 859003 51261 40866 592510 556235 324926 700261 320946 798...
output:
0 0 953730 0 0 859562 0 0 0 0 776333 0 0 0 0 883717 0 0 0 0 0 0 899232 0 0 0 0 0 0 0 0 0 0 0 896479 0 0 892545 0 0 0 0 907449 0 0 0 892956 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 898541 788810 0 0 0 0 0 0 891526 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 903499 0 0 0 0 0 0 0 914433 0 0 ...
result:
wrong answer 6th numbers differ - expected: '512663', found: '859562'
Subtask #3:
score: 0
Wrong Answer
Test #17:
score: 0
Wrong Answer
time: 2932ms
memory: 125220kb
input:
999995 999997 261379 334440 985281 986034 549380 718157 670003 253925 533027 437989 326806 983213 965935 756259 229069 686789 331338 684961 957559 390618 937820 959719 338153 779814 582581 965418 634156 421264 308778 938878 913059 390904 481477 431492 739774 793015 901442 934600 256704 991485 366691...
output:
160637 633545 123521 313579 450015 383638 574188 2378 203701 326075 117994 567277 652826 199783 380319 380379 105882 373861 788346 703180 609805 648165 367068 497984 57990 478851 745058 75946 543070 251847 769828 436497 480976 235845 509633 757192 763494 396852 262690 2262 267186 165800 558973 16525...
result:
wrong answer 1st numbers differ - expected: '160635', found: '160637'
Subtask #4:
score: 0
Skipped
Dependency #1:
0%
Subtask #5:
score: 0
Skipped
Dependency #1:
0%