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ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#254128 | #7680. Subway | ucup-team134# | WA | 1ms | 4148kb | C++17 | 4.5kb | 2023-11-18 01:21:23 | 2023-11-18 01:21:24 |
Judging History
answer
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/rope>
#define ll long long
#define pb push_back
#define f first
#define s second
#define sz(x) (int)(x).size()
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define ios ios_base::sync_with_stdio(false);cin.tie(NULL)
#define ld long double
#define li __int128
using namespace std;
using namespace __gnu_pbds;
using namespace __gnu_cxx;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>; ///find_by_order(),order_of_key()
template<int D, typename T>struct vec : public vector<vec<D - 1, T>> {template<typename... Args>vec(int n = 0, Args... args) : vector<vec<D - 1, T>>(n, vec<D - 1, T>(args...)) {}};
template<typename T>struct vec<1, T> : public vector<T> {vec(int n = 0, T val = T()) : vector<T>(n, val) {}};
template<class T1,class T2> ostream& operator<<(ostream& os, const pair<T1,T2>& a) { os << '{' << a.f << ", " << a.s << '}'; return os; }
template<class T> ostream& operator<<(ostream& os, const vector<T>& a){os << '{';for(int i=0;i<sz(a);i++){if(i>0&&i<sz(a))os << ", ";os << a[i];}os<<'}';return os;}
template<class T> ostream& operator<<(ostream& os, const deque<T>& a){os << '{';for(int i=0;i<sz(a);i++){if(i>0&&i<sz(a))os << ", ";os << a[i];}os<<'}';return os;}
template<class T> ostream& operator<<(ostream& os, const set<T>& a) {os << '{';int i=0;for(auto p:a){if(i>0&&i<sz(a))os << ", ";os << p;i++;}os << '}';return os;}
template<class T> ostream& operator<<(ostream& os, const set<T,greater<T> >& a) {os << '{';int i=0;for(auto p:a){if(i>0&&i<sz(a))os << ", ";os << p;i++;}os << '}';return os;}
template<class T> ostream& operator<<(ostream& os, const multiset<T>& a) {os << '{';int i=0;for(auto p:a){if(i>0&&i<sz(a))os << ", ";os << p;i++;}os << '}';return os;}
template<class T> ostream& operator<<(ostream& os, const multiset<T,greater<T> >& a) {os << '{';int i=0;for(auto p:a){if(i>0&&i<sz(a))os << ", ";os << p;i++;}os << '}';return os;}
template<class T1,class T2> ostream& operator<<(ostream& os, const map<T1,T2>& a) {os << '{';int i=0;for(auto p:a){if(i>0&&i<sz(a))os << ", ";os << p;i++;}os << '}';return os;}
int ri(){int x;scanf("%i",&x);return x;}
void rd(int&x){scanf("%i",&x);}
void rd(long long&x){scanf("%lld",&x);}
void rd(double&x){scanf("%lf",&x);}
void rd(long double&x){scanf("%Lf",&x);}
void rd(string&x){cin>>x;}
void rd(char*x){scanf("%s",x);}
template<typename T1,typename T2>void rd(pair<T1,T2>&x){rd(x.first);rd(x.second);}
template<typename T>void rd(vector<T>&x){for(T&p:x)rd(p);}
template<typename C,typename...T>void rd(C&a,T&...args){rd(a);rd(args...);}
//istream& operator>>(istream& is,__int128& a){string s;is>>s;a=0;int i=0;bool neg=false;if(s[0]=='-')neg=true,i++;for(;i<s.size();i++)a=a*10+s[i]-'0';if(neg)a*=-1;return is;}
//ostream& operator<<(ostream& os,__int128 a){bool neg=false;if(a<0)neg=true,a*=-1;ll high=(a/(__int128)1e18);ll low=(a-(__int128)1e18*high);string res;if(neg)res+='-';if(high>0){res+=to_string(high);string temp=to_string(low);res+=string(18-temp.size(),'0');res+=temp;}else res+=to_string(low);os<<res;return os;}
int main()
{
int n=ri();
vector<pair<pair<int,int>,int>> ok;
int l=0;
for(int i=0;i<n;i++){
int x=ri(),y=ri(),a=ri();
ok.pb({{y,x},a});
l=max(l,a);
}
sort(all(ok));
int X=1e8;
int Y=-1001;
vector<vector<pair<int,int>>> lines(l);
vector<int> lastStation(l,-1);
vector<int> sty;
vector<int> stx;
for(int i=0;i<n;i++){
int x=ok[i].f.s,y=ok[i].f.f;
int a=ok[i].s;
sty.pb(Y-1);
int stX=X-1;
stx.pb(stX);
for(int j=0;j<a;j++){
int lind=j;
if(lastStation[lind]!=-1){ //jump!
if(lastStation[lind]==i-1){
lines[lind].pb({stx[i],sty[i]});
sty[i]--;
}
else{
int yy=0;
for(int k=i;k>lastStation[lind];k--){
yy=min(yy,sty[k]);
}
for(int k=i;k>lastStation[lind];k--){
sty[k]=yy+1;
}
lines[lind].pb({stx[lastStation[lind]+1],yy});
lines[lind].pb({stx[i],yy});
}
}
lines[lind].pb({X,Y});
lines[lind].pb({x,y});
X+=2;
}
for(int j=0;j<a;j++){
int lind=a-j-1;
lines[lind].pb({X,Y});
X+=2;
lastStation[lind]=i;
}
}
printf("%i\n",l);
for(int i=0;i<l;i++){
printf("%i",sz(lines[i]));
for(auto p:lines[i])printf(" %i %i",p.f,p.s);
printf("\n");
}
return 0;
}
详细
Test #1:
score: 100
Accepted
time: 0ms
memory: 3872kb
input:
3 1 2 1 2 1 2 3 3 2
output:
2 11 100000000 -1001 2 1 100000006 -1001 100000007 -1002 100000008 -1001 1 2 100000010 -1001 100000011 -1002 100000012 -1001 3 3 100000018 -1001 8 100000002 -1001 2 1 100000004 -1001 100000007 -1003 100000011 -1003 100000014 -1001 3 3 100000016 -1001
result:
ok ok Sum L = 19
Test #2:
score: 0
Accepted
time: 0ms
memory: 3860kb
input:
1 1 1 1
output:
1 3 100000000 -1001 1 1 100000002 -1001
result:
ok ok Sum L = 3
Test #3:
score: 0
Accepted
time: 0ms
memory: 4036kb
input:
1 1 1 50
output:
50 3 100000000 -1001 1 1 100000198 -1001 3 100000002 -1001 1 1 100000196 -1001 3 100000004 -1001 1 1 100000194 -1001 3 100000006 -1001 1 1 100000192 -1001 3 100000008 -1001 1 1 100000190 -1001 3 100000010 -1001 1 1 100000188 -1001 3 100000012 -1001 1 1 100000186 -1001 3 100000014 -1001 1 1 100000184...
result:
ok ok Sum L = 150
Test #4:
score: -100
Wrong Answer
time: 1ms
memory: 4148kb
input:
50 662 -567 48 728 -120 7 307 669 27 -885 -775 21 100 242 9 -784 -537 41 940 198 46 736 -551 30 -449 456 16 -945 382 18 -182 810 49 213 187 44 853 245 48 617 -305 19 -81 261 3 617 208 8 -548 -652 6 -888 -667 14 -371 -812 43 202 -702 10 -668 -725 5 961 -919 33 -870 -697 50 428 810 29 560 405 7 348 -3...
output:
50 199 100000000 -1001 961 -919 100000130 -1001 100000131 -1002 100000132 -1001 -306 -897 100000302 -1001 100000303 -1002 100000304 -1001 334 -893 100000334 -1001 100000335 -1002 100000336 -1001 -371 -812 100000506 -1001 100000507 -1002 100000508 -1001 -885 -775 100000590 -1001 100000591 -1002 10000...
result:
wrong answer Polyline 3 intersects with previous polylines.