이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
/* There's someone in my head but it's not me */
#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef pair<int, int> pii;
#define SZ(x) (int) x.size()
#define F first
#define S second
const int N = 3e5 + 10, MOD = 1e9 + 7;
int n, X[N], Y[N], R[N], ord[N], ret[N];
int intersect(int i, int j) {
ll x1 = X[i], x2 = X[j], y1 = Y[i], y2 = Y[j], r1 = R[i], r2 = R[j];
return x1 * x1 + x2 * x2 + y1 * y1 + y2 * y2 - r1 * r1 - r2 * r2 <= 2 * (r1 * r2 + x1 * x2 + y1 * y2);
}
void sub1() {
for (int _ = 1; _ <= n; _++) {
int i = ord[_];
if (ret[i]) continue;
ret[i] = i;
for (int j = _ + 1; j <= n; j++) {
if (!ret[ord[j]] && intersect(i, ord[j])) ret[ord[j]] = i;
}
}
for (int i = 1; i <= n; i++)
printf("%d ", ret[i]);
printf("\n");
exit(0);
}
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; i++)
scanf("%d%d%d", &X[i], &Y[i], &R[i]);
iota(ord + 1, ord + n + 1, 1);
sort(ord + 1, ord + n + 1, [&] (int i, int j) { return R[i] == R[j] ? i < j : R[i] > R[j]; });
if (1ll * n * n <= 1e8) sub1();
set<pair<ll, int>, greater<pair<ll, int>>> st;
for (int i = 1; i <= n; i++)
st.insert({X[i] + R[i], i});
for (int _ = 1; _ <= n; _++) {
int i = ord[_];
if (ret[i]) continue;
ret[i] = i; st.erase({X[i] + R[i], i});
while (SZ(st) && st.begin()->F >= X[i] - R[i]) {
ret[st.begin()->S] = i;
st.erase(st.begin());
}
}
for (int i = 1; i <= n; i++) printf("%d ", ret[i]);
printf("\n");
return 0;
}
컴파일 시 표준 에러 (stderr) 메시지
circle_selection.cpp: In function 'int main()':
circle_selection.cpp:37:10: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
37 | scanf("%d", &n);
| ~~~~~^~~~~~~~~~
circle_selection.cpp:39:14: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
39 | scanf("%d%d%d", &X[i], &Y[i], &R[i]);
| ~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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