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QOJ
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 |
---|---|---|---|---|---|---|---|---|---|
#136599 | #237. Triangle Partition | 4k2kok# | 100 ✓ | 8ms | 3608kb | C++20 | 1005b | 2023-08-09 09:15:25 | 2023-08-09 09:15:29 |
Judging History
answer
#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define io ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define mem(a,b) memset((a),(b),sizeof(a))
typedef long long ll;
typedef unsigned long long ull;
#define int long long
#define db long double
const double eps = 1e-6;
const int mod = 1e9 + 7;
const int maxn = 1e5 + 10;
int n;
struct node{
int x, y, id;
bool operator < (const node&A) const{
if(x == A.x) return y < A.y;
else return x < A.x;
}
};
void solve() {
cin >> n;
vector<node> vec;
for(int i = 1, u, v; i <= 3 * n; i++) {
cin >> u >> v;
vec.push_back({u, v, i});
}
sort(vec.begin(), vec.end());
for(int i = 0; i < 3 * n; i += 3) {
cout << vec[i].id << ' ' << vec[i + 1].id << ' ' << vec[i + 2].id << '\n';
}
}
signed main(){
io;
int T = 1;
cin >> T;
while(T--) {
solve();
}
return 0;
}
詳細信息
Test #1:
score: 100
Accepted
time: 8ms
memory: 3608kb
input:
190 10 -7215 2904 -5179 1663 -542 1091 -5687 7868 7838 -1048 -2944 4346 -2780 3959 -9402 1099 -8548 -7238 -3821 -2917 2713 295 -856 -8661 7651 3945 -8216 -543 5798 5024 8583 -3384 -1208 5955 3068 -385 340 2968 6559 -272 4537 5075 5126 4343 639 8281 1700 2572 819 9317 -9854 -1316 -3421 -1137 9368 718...
output:
26 8 9 14 29 1 4 2 10 27 6 7 17 12 3 19 23 25 24 11 18 21 22 15 20 30 13 5 16 28 30 3 18 29 5 17 12 11 27 15 19 10 28 25 9 24 7 13 6 8 23 14 22 2 20 4 16 26 21 1 6 11 16 8 27 10 13 15 14 29 22 19 9 2 18 24 12 20 1 26 4 5 30 7 3 25 23 17 21 28 16 8 18 19 23 13 7 26 24 3 1 5 29 25 27 9 6 15 12 22 14 1...
result:
ok AC