QOJ.ac

QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#123551#6639. Disk TreeinstallbWA 2ms9548kbC++142.2kb2023-07-12 20:37:542023-07-12 20:37:56

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2023-07-12 20:37:56]
  • 评测
  • 测评结果:WA
  • 用时:2ms
  • 内存:9548kb
  • [2023-07-12 20:37:54]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 400005;

struct interval{
    int l,r,id;
    bool operator < (const interval &o)const{
        return l < o.l;
    }
}s[N];

set <pair <int,int> > st;
vector <tuple <int,int,int,int> > ans;
int n,x[N],y[N],r[N];

void solve(){
    cin >> n;
    for(int i = 1;i <= n;i ++){
        cin >> x[i] >> y[i] >> r[i];
        s[i] = {x[i] - r[i],x[i] + r[i],i};
    }
    sort(s + 1,s + 1 + n);
    // x has to be >= 0, so connect all possible circles with x = 0 first
    int pos = 0;
    for(int i = 1;i <= n;i ++){
        if(x[s[i].id] - r[s[i].id] <= 0) pos = i;
        else break;
    }
    vector <pair <int,int> > G;
    for(int i = 1;i <= pos;i ++){
        G.push_back({y[s[i].id],s[i].id});
        st.insert({y[s[i].id],s[i].id});
    }
    sort(G.begin(),G.end());
    for(int i = 1;i < G.size();i ++){
        int u = G[i - 1].second,v = G[i].second;
        ans.push_back({0,y[u],0,y[v]});
    }
    
    pair <int,int> mxr = {-1,-1};
    for(int i = 1;i <= pos;i ++) mxr = max(mxr,{x[s[i].id] + r[s[i].id],s[i].id});
    for(int i = pos + 1;i <= n;i ++){
        int u = s[i].id,id = -1;
        while(st.size()){
            auto it = st.lower_bound({y[u],-1});
            if(it == st.end()) break;
            int v = it->second;
            if(x[v] + r[v] < x[u] - r[u]) st.erase(it);
            else{
                id = v;
                break;
            }
        }
        if(id == -1){
            while(st.size()){
                auto it = -- st.lower_bound({y[u],-1});
                int v = it->second;
                if(x[v] + r[v] < x[u] - r[u]) st.erase(it);
                else{
                    id = v;
                    break;
                }
            }
        }
        if(id == -1 && i > 1) ans.push_back({mxr.first,y[mxr.second],x[u] - r[u],y[u]});
        else ans.push_back({x[u] - r[u],y[id],x[u] - r[u],y[u]});
        st.insert({y[u],u});
        mxr = max(mxr,{x[u] + r[u],u});
    }

    cout << "YES\n";
    for(auto [a,b,c,d] : ans) cout << a << ' ' << b << ' ' << c << ' ' << d << '\n';
}

int main(){
    ios::sync_with_stdio(false);
    solve();
    return 0;
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 2ms
memory: 7504kb

input:

3
1 0 3
10 10 6
0 5 1

output:

YES
0 0 0 5
4 0 4 10

result:

ok answer = 1

Test #2:

score: 0
Accepted
time: 1ms
memory: 7500kb

input:

2
1 1 1
3 3 1

output:

YES
2 1 2 3

result:

ok answer = 1

Test #3:

score: 0
Accepted
time: 1ms
memory: 9548kb

input:

5
10 10 10
2 0 1
20 20 1
3 20 1
20 0 1

output:

YES
1 10 1 0
2 10 2 20
19 10 19 20
19 10 19 0

result:

ok answer = 1

Test #4:

score: -100
Wrong Answer
time: 1ms
memory: 7580kb

input:

10
29 29 2
28 55 10
99 81 4
17 82 10
45 88 10
48 68 10
0 8 10
98 95 10
34 0 10
17 24 10

output:

YES
7 8 7 82
7 82 7 24
18 82 18 55
24 24 24 0
27 55 27 29
35 55 35 88
38 88 38 68
58 68 88 95
95 95 95 81

result:

wrong answer Two line segments intersect, and it's not only the endpoints that intersect or line segments intersects/touches more than 2 disks