提出 #58211442

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ソースコード 拡げる

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef pair<ll, ll> pll;

// constants
const bool HAVE_CASE = false;
const ll MOD = 998244353;

// A C++ program to check if two given line segments intersect 
#include <iostream> 
using namespace std; 
  
struct Point 
{ 
    int x; 
    int y; 
}; 
  
// Given three collinear points p, q, r, the function checks if 
// point q lies on line segment 'pr' 
bool onSegment(Point p, Point q, Point r) 
{ 
    if (q.x <= max(p.x, r.x) && q.x >= min(p.x, r.x) && 
        q.y <= max(p.y, r.y) && q.y >= min(p.y, r.y)) 
       return true; 
  
    return false; 
} 
  
// To find orientation of ordered triplet (p, q, r). 
// The function returns following values 
// 0 --> p, q and r are collinear 
// 1 --> Clockwise 
// 2 --> Counterclockwise 
int orientation(Point p, Point q, Point r) 
{ 
    // See https://www.geeksforgeeks.org/orientation-3-ordered-points/ 
    // for details of below formula. 
    int val = (q.y - p.y) * (r.x - q.x) - 
              (q.x - p.x) * (r.y - q.y); 
  
    if (val == 0) return 0;  // collinear 
  
    return (val > 0)? 1: 2; // clock or counterclock wise 
} 
  
// The main function that returns true if line segment 'p1q1' 
// and 'p2q2' intersect. 
bool doIntersect(Point p1, Point q1, Point p2, Point q2) 
{ 
    // Find the four orientations needed for general and 
    // special cases 
    int o1 = orientation(p1, q1, p2); 
    int o2 = orientation(p1, q1, q2); 
    int o3 = orientation(p2, q2, p1); 
    int o4 = orientation(p2, q2, q1); 
  
    // General case 
    if (o1 != o2 && o3 != o4) 
        return true; 
  
    // Special Cases 
    // p1, q1 and p2 are collinear and p2 lies on segment p1q1 
    if (o1 == 0 && onSegment(p1, p2, q1)) return true; 
  
    // p1, q1 and q2 are collinear and q2 lies on segment p1q1 
    if (o2 == 0 && onSegment(p1, q2, q1)) return true; 
  
    // p2, q2 and p1 are collinear and p1 lies on segment p2q2 
    if (o3 == 0 && onSegment(p2, p1, q2)) return true; 
  
     // p2, q2 and q1 are collinear and q1 lies on segment p2q2 
    if (o4 == 0 && onSegment(p2, q1, q2)) return true; 
  
    return false; // Doesn't fall in any of the above cases 
}

void test_case_run() {
    int n;
    cin >> n;
    Point P[n+1], Q[n+1];
    for (int i = 1; i <= n; i++) cin >> P[i].x >> P[i].y;
    for (int i = 1; i <= n; i++) cin >> Q[i].x >> Q[i].y;
    int R[n+1];
    iota(R, R+n+1, 0);

    bool pass = false;
    for (int itr = 1; !pass && itr <= 2*n; itr++) {
        pass = true;
        for (int i = 1; i <= n; i++) {
            for (int j = i+1; j <= n; j++) {
                if (doIntersect(P[i], Q[R[i]], P[j], Q[R[j]])) {
                    pass = false;
                    swap(R[i], R[j]);
                }
            }
        }
    }

    if (!pass) cout << -1;
    else for (int i = 1; i <= n; i++) cout << R[i] << " ";
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    ll t = 1;
    if (HAVE_CASE) cin >> t;
    while (t--) test_case_run();

    cout.flush();
    
    return 0;
}

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