QOJ.ac

QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#59300#4418. Laser AlarmqinjianbinAC ✓146ms3760kbC++5.2kb2022-10-29 10:38:532022-10-29 10:38:57

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.
  • [2022-10-29 10:38:57]
  • 评测
  • 测评结果:AC
  • 用时:146ms
  • 内存:3760kb
  • [2022-10-29 10:38:53]
  • 提交

answer

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
#include<cctype>
#include<ctime>
#include<queue>
#include<stack>
#include<deque>
#include<map>
#include<unordered_map>
#include<set>
#include<unordered_set>
#include<bitset>
#include<functional>
using namespace std;
//STL库常用(vector, map, set, pair)
#define pii pair<int,int>
#define pdd pair<double,double>
#define F first
#define S second
//常量
#define PI (acos(-1.0))
#define INF 0x3f3f3f3f
//必加
#ifdef TanJI
#include<cassert>
#define dbg(...) do {cerr << "\033[33;1m" << #__VA_ARGS__ << " -> "; err(__VA_ARGS__); } while(0)
void err() {cerr << "\033[39;0m" << endl;}
template<template<typename...> class T, typename t, typename... A> void err(T<t> a, A... x) {for (auto v: a) cerr << v << ' '; cerr << ", "; err(x...);}
template<typename T, typename... A> void err(T a, A... x) {cerr << a << ' '; err(x...);}
template<typename T> void err(T *a, int len) {for (int i = 0; i < len; i++) cerr << a[i] << ' '; err();}
#else 
#define dbg(...)
#define assert(...)
#endif
typedef unsigned long long ull;
typedef long long ll;
typedef double db;
// 快读快输
template<typename T> inline T read() {T x=0; bool f=0; char c=getchar();while(!isdigit(c)){if(c =='-')f= 1; c=getchar();}while(isdigit(c)){x=(x<<3)+(x<<1)+(c^48);c=getchar();}return f?~x+1:x;}
template<typename T> inline void print(T k){ int num = 0,ch[20]; if(k == 0){ putchar('0'); return ; } (k<0)&&(putchar('-'),k = -k); while(k>0) ch[++num] = k%10, k /= 10; while(num) putchar(ch[num--]+48); }
//图论
const int GRAPHN = 1, GRAPHM = 1;
int h[GRAPHN], e[GRAPHM], ne[GRAPHM], wei[GRAPHM], idx;
void AddEdge(int a, int b) {e[idx] = b, ne[idx] = h[a], h[a] = idx++;}
void AddEdge(int a, int b, int c) {e[idx] = b, ne[idx] = h[a], wei[idx] = c, h[a] = idx++;}
void AddEdge(int h[], int a, int b) {e[idx] = b, ne[idx] = h[a], h[a] = idx++;}
void AddEdge(int h[], int a, int b, int c) {e[idx] = b, ne[idx] = h[a], wei[idx] = c, h[a] = idx++;}
// 数学公式
template<typename T> inline T gcd(T a, T b){ return b==0 ? a : gcd(b,a%b); }
template<typename T> inline T lowbit(T x){ return x&(-x); }
template<typename T> inline bool mishu(T x){ return x>0?(x&(x-1))==0:false; }
template<typename T1,typename T2, typename T3> inline ll q_mul(T1 a,T2 b,T3 p){ ll w = 0; while(b){ if(b&1) w = (w+a)%p; b>>=1; a = (a+a)%p; } return w; }
template<typename T,typename T2> inline ll f_mul(T a,T b,T2 p){ return (a*b - (ll)((long double)a/p*b)*p+p)%p; }
template<typename T1,typename T2, typename T3> inline ll q_pow(T1 a,T2 b,T3 p){ ll w = 1; while(b){ if(b&1) w = (w*a)%p; b>>=1; a = (a*a)%p;} return w; }
template<typename T1,typename T2, typename T3> inline ll s_pow(T1 a,T2 b,T3 p){ ll w = 1; while(b){ if(b&1) w = q_mul(w,a,p); b>>=1; a = q_mul(a,a,p);} return w; }
template<typename T> inline ll ex_gcd(T a, T b, T& x, T& y){ if(b == 0){ x = 1, y = 0; return (ll)a; } ll r = ex_gcd(b,a%b,y,x); y -= a/b*x; return r;/*gcd*/ }
template<typename T1,typename T2> inline ll com(T1 m, T2 n) { int k = 1;ll ans = 1; while(k <= n){ ans=((m-k+1)*ans)/k;k++;} return ans; }
template<typename T> inline bool isprime(T n){ if(n <= 3) return n>1; if(n%6 != 1 && n%6 != 5) return 0; T n_s = floor(sqrt((db)(n))); for(int i = 5; i <= n_s; i += 6){ if(n%i == 0 || n%(i+2) == 0) return 0; } return 1; }
/* ----------------------------------------------------------------------------------------------------------------------------------------------------------------- */

const int MAXN = 110;
struct Point {
    int x, y, z;
}p[MAXN];
Point Cross(Point a, Point b) {
    return {
         a.y * b.z - a.z * b.y,
         a.z * b.x - a.x * b.z,
         a.x * b.y - a.y * b.x
    };
}
int Dot(Point a, Point b) {
    return a.x * b.x + a.y * b.y + a.z * b.z;
}
Point GetN(Point a, Point b, Point c) {
    return Cross({b.x - a.x, b.y - a.y, b.z - a.z}, {c.x - a.x, c.y - a.y, c.z - a.z});
}
Point Vector(Point sta, Point ed) {
    return {ed.x - sta.x, ed.y - sta.y, ed.z - sta.z};
}
int T, n;


int main() {
#ifdef TanJI
    clock_t c1 = clock();
    freopen("D:\\Cpp\\1.in", "r", stdin);
    freopen("D:\\Cpp\\1.out", "w", stdout);
#endif
//--------------------------------------------
    scanf("%d", &T);
    while (T--) {
        scanf("%d", &n);
        n <<= 1;
        for (int i = 1; i <= n; i++)
            scanf("%d%d%d", &p[i].x, &p[i].y, &p[i].z);
        int ans = 0;
        for (int i = 1; i <= n; i++) {
            for (int j = i + 1; j <= n; j++) {
                for (int k = j + 1; k <= n; k++) {
                    int res = 0;
                    Point nn = GetN(p[i], p[j], p[k]);
                    if (nn.x == 0 && nn.y == 0 && nn.z == 0) continue;
                    for (int z = 1; z <= n; z += 2) {
                        if (1ll * Dot(nn, Vector(p[i], p[z])) * Dot(nn, Vector(p[i], p[z + 1])) <= 0) 
                            res++;
                    }
                    ans = max(ans, res);
                }
            }
        }
        if (ans == 0) ans = n / 2;
        printf("%d\n", ans);
    }
//--------------------------------------------    
#ifdef TanJI
    cerr << "Time Used:" << clock() - c1 << "ms" << endl;
#endif
    return 0;
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 146ms
memory: 3760kb

input:

10
50
94 57 39 12 69 59
86 46 44 17 32 83
35 86 71 47 41 50
68 93 71 54 28 25
92 74 2 30 60 86
87 52 54 32 17 88
51 63 96 23 12 69
1 82 85 20 9 90
25 72 42 49 4 52
30 86 94 93 43 34
10 45 30 85 32 75
84 37 71 37 78 19
28 30 7 40 10 77
5 68 86 83 3 41
71 73 8 86 69 48
65 11 6 49 64 50
61 2 24 60 11 9...

output:

33
39
36
36
40
37
34
33
50
50

result:

ok 10 lines