제출 #254777

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
254777		MarcoMeijer	Triangles (CEOI18_tri)	C++14	100 / 100	48 ms	4344 KiB

tri

#include <bits/stdc++.h>
#include "trilib.h"
using namespace std;
 
// macros
typedef long long ll;
typedef long double ld;
typedef pair<int, int> ii;
typedef pair<ll, ll> lll;
typedef tuple<int, int, int> iii;
typedef vector<int> vi;
typedef vector<ii> vii;
typedef vector<iii> viii;
typedef vector<ll> vll;
typedef vector<lll> vlll;
#define REP(a,b,c) for(int a=int(b); a<int(c); a++)
#define RE(a,c) REP(a,0,c)
#define RE1(a,c) REP(a,1,c+1)
#define REI(a,b,c) REP(a,b,c+1)
#define REV(a,b,c) for(int a=int(c-1); a>=int(b); a--)
#define FOR(a,b) for(auto& a : b)
#define all(a) a.begin(), a.end()
#define INF 1e9
#define EPS 1e-9
#define pb push_back
#define popb pop_back
#define fi first
#define se second
#define sz size()
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
 
// input
template<class T> void IN(T& x) {cin >> x;}
template<class H, class... T> void IN(H& h, T&... t) {IN(h); IN(t...); }
 
// output
template<class T1, class T2> void OUT(const pair<T1,T2>& x);
template<class T> void OUT(const T& x) {cout << x;}
template<class H, class... T> void OUT(const H& h, const T&... t) {OUT(h); OUT(t...); }
template<class T1, class T2> void OUT(const pair<T1,T2>& x) {OUT(x.fi," ",x.se);}
template<class... T> void OUTL(const T&... t) {OUT(t..., "\n"); }
 
//===================//
//  Added libraries  //
//===================//
 
//===================//
//end added libraries//
//===================//
 
void program();
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    program();
}
 
 
//===================//
//   begin program   //
//===================//
 
const int MX = 50000;
 
int n, m;
int SHUF[MX];
bool cw(int x, int y, int z) {
    return is_clockwise(SHUF[x],SHUF[y],SHUF[z]);
}
bool ccw(int x, int y, int z) {
    return !is_clockwise(SHUF[x],SHUF[y],SHUF[z]);
}
int MEX(int x, int y) {
    RE1(i,3)
        if(x != i && y != i) return i;
}
 
int origin;
int S[2][MX], C[2][MX], N[2], M[2];
set<int> onSide[2];
int ST[2][MX], O[2];
int s1, s2;
 
void program() {
    n = get_n();
    RE1(i,n) SHUF[i] = i;
    shuffle(SHUF+1,SHUF+n+1,rng);
    if(n == 3) {
        give_answer(3);
        return;
    }
 
    // create two halves
    s1 = 1;
    s2 = 2;
    REI(j,3,n) onSide[cw(s1,s2,j)].insert(j);
    RE(i,2) {
        // i == 0 left side of 1->2, i == 1 right side of 1->2
        onSide[i].insert(s1);
        onSide[i].insert(s2);
        origin = i+1;
        N[i] = 0;
        FOR(j,onSide[i]) S[i][N[i]++] = j;
        RE(j,N[i]) if(S[i][j] == origin) swap(S[i][j], S[i][0]);
        sort(S[i]+1,S[i]+N[i],[](int i, int j) {
            return ccw(origin, i, j);
        });
    
        M[i]=0;
        C[i][M[i]++] = S[i][N[i]-1], C[i][M[i]++] = S[i][0];
        int j=1;
        if(N[i] != 2) {
            while(j < N[i]) {
                if(ccw(C[i][M[i]-2],C[i][M[i]-1],S[i][j])) C[i][M[i]++] = S[i][j++];
                else M[i]--;
            }
        } else {
            C[i][M[i]++] = S[i][1];
        }
    }

    // merge two halves
    int answer = M[0]+M[1]-2-2;
    RE(h,2) {
        RE(i,2) O[i]=0;
        REP(j,2,M[ h]) ST[ h][O[ h]++] = C[ h][j];
        REP(j,2,M[!h]) ST[!h][O[!h]++] = C[!h][j];
        reverse(ST[h], ST[h]+O[h]);
        int c[2];
        RE(i,2) c[i]=0;
        while((O[h]-c[h]) && (O[!h]-c[!h])) {
            if(O[h]-c[h] >= 2) {
                if(cw(ST[h][c[h]+1],ST[h][c[h]],ST[!h][c[!h]])) {
                    answer--;
                    c[h]++;
                    continue;
                }
            }
            if(O[!h]-c[!h] >= 2) {
                if(ccw(ST[!h][c[!h]+1],ST[!h][c[!h]],ST[h][c[h]])) {
                    answer--;
                    c[!h]++;
                    continue;
                }
            }
            break;
        }
    }

    give_answer(answer);
}

컴파일 시 표준 에러 (stderr) 메시지

tri.cpp: In function 'int MEX(int, int)':
tri.cpp:77:1: warning: control reaches end of non-void function [-Wreturn-type]
 }
 ^

• Subtask 1: 15 / 15
• Subtask 2: 20 / 20
• Subtask 3: 20 / 20
• Subtask 4: 20 / 20
• Subtask 5: 25 / 25