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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
#573393#7906. Almost Convexrxzfn639TL 0ms4160kbC++233.9kb2024-09-18 18:32:502024-09-18 18:32:51

Judging History

你现在查看的是最新测评结果

  • [2024-09-18 18:32:51]
  • 评测
  • 测评结果:TL
  • 用时:0ms
  • 内存:4160kb
  • [2024-09-18 18:32:50]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'
#define il inline
const int N = 4e3 + 5;


const double eps = 1e-8;
il int sgn(double x)
{
    // 进行判断, 提高精度
    if (fabs(x) <= eps)
        return 0;          // x == 0, 精度范围内的近似相等
    return x > 0 ? 1 : -1; // 返回正负
}
il bool eq(double a, double b) { return abs(a - b) < eps; } // ==
il bool le(double a, double b) { return a - b < eps; }      // <=
il bool lt(double a, double b) { return a - b < -eps; }     // <
typedef struct Point
{
    int x, y;
    // Point(double x = 0, double y = 0) : x(x), y(y) {} // 构造函数, 初始值为 0

    // 重载运算符
    // 点 - 点 = 向量 (向量AB = B - A)
    Point operator-(const Point &B) const { return Point(x - B.x, y - B.y); }

    // 点 + 点 = 点, 点 + 向量 = 向量, 向量 + 向量 = 向量
    Point operator+(const Point &B) const { return Point(x + B.x, y + B.y); }

    // 向量 · 向量 (点积)
    double operator*(const Point &B) const { return x * B.x + y * B.y; }

    // 向量 × 向量 (叉积)
    double operator^(const Point &B) const { return x * B.y - y * B.x; }

    // 判断大小, 一般用于排序
    bool operator<(const Point &B) const { return x < B.x || (x == B.x && y < B.y); }

} Vector;

using Points = vector<Point>;

// 极角排序
// Need: (^, sgn)
// 基准点
Point p0;
il double theta(auto p) { return atan2(p.y, p.x); } // 求极角
void psort(Points &ps, Point c = p0)              // 极角排序
{
    sort(ps.begin(), ps.end(), [&](auto p1, auto p2) {
        return lt(theta(p1 - c), theta(p2 - c));
    });
}

// 凸包
// Andrew算法求凸包,最后一个点与第一个点重合
// Need: (^,-,<), sgn, le
il bool check(Point p, Point q, Point r) { return le(0, (q - p) ^ (r - q)); }
vector<Point> Andrew(Points poly)
{
    int n = poly.size(), top = 0;
    vector<int> stk(n + 10, 0), used(n + 10, 0);
    sort(poly.begin(), poly.end());
    stk[++top] = 0;
    for (int i = 1; i < n; i++)
    {
        while (top > 1 && sgn((poly[stk[top]] - poly[stk[top - 1]]) ^ (poly[i] - poly[stk[top]])) <= 0)
            used[stk[top--]] = 0;

        used[i] = 1;
        stk[++top] = i;
    }
    int tmp = top;
    for (int i = n - 2; i >= 0; i--)
    {
        if (used[i]) continue;
        while (top > tmp && sgn((poly[stk[top]] - poly[stk[top - 1]]) ^ (poly[i] - poly[stk[top]])) <= 0)
            used[stk[top--]] = 0;

        used[i] = 1;
        stk[++top] = i;
    }
    vector<Point> a;
    for (int i = 1; i <= top; i++) a.push_back(poly[stk[i]]);
    return a;
}

const int mod = 1e9 + 7;
int hs(int x, int y)
{
    return (x * 100000000 + y) % mod;
}
Point temp[N];
void solve()
{   int o = 0;
    int n;
    cin >> n;
    Points ans, p;
    for (int i = 0; i < n; i++)
    {
        int a, b;
        cin >> a >> b;
        p.push_back({a, b});
    }
    ans = Andrew(p);

    unordered_map<int, int> mp;
    for (auto [x, y]: ans) mp[hs(x, y)] = 1;
    
    int res = 0;
    for (int i = 0; i < n; i++)
    {
        auto [x, y] = p[i];
        if (mp[hs(x, y)]) continue;

        p0 = p[i];
        int tn = 0;
        for (int j = 0; j < n; j++) 
        {
            o++;
            if (j != i) temp[tn++] = p[j];
        }
        sort(temp, temp + tn, [&](auto p1, auto p2) {
            return lt(theta(p1 - p0), theta(p2 - p0));
        });

        for (int j = 0; j < tn; j++)
        {
            o++;
            auto p1 = temp[j], p2 = temp[(j + 1) % tn];
            if(mp[hs(p1.x, p1.y)] && mp[hs(p2.x, p2.y)]) res++;
        }
        if (n == 2000) cout << o << '\n';
    }
    cout << res + 1 << endl;
}
signed main()
{
    // ios::sync_with_stdio(false);
    // cin.tie(0); cout.tie(0);
    int T = 1; 
    while (T--) solve();
    return 0;
}

详细

Test #1:

score: 100
Accepted
time: 0ms
memory: 3992kb

input:

7
1 4
4 0
2 3
3 1
3 5
0 0
2 4

output:

9

result:

ok 1 number(s): "9"

Test #2:

score: 0
Accepted
time: 0ms
memory: 4064kb

input:

5
4 0
0 0
2 1
3 3
3 1

output:

5

result:

ok 1 number(s): "5"

Test #3:

score: 0
Accepted
time: 0ms
memory: 3496kb

input:

3
0 0
3 0
0 3

output:

1

result:

ok 1 number(s): "1"

Test #4:

score: 0
Accepted
time: 0ms
memory: 4160kb

input:

6
0 0
3 0
3 2
0 2
1 1
2 1

output:

7

result:

ok 1 number(s): "7"

Test #5:

score: 0
Accepted
time: 0ms
memory: 3620kb

input:

4
0 0
0 3
3 0
3 3

output:

1

result:

ok 1 number(s): "1"

Test #6:

score: -100
Time Limit Exceeded

input:

2000
86166 617851
383354 -277127
844986 386868
-577988 453392
-341125 -386775
-543914 -210860
-429613 606701
-343534 893727
841399 339305
446761 -327040
-218558 -907983
787284 361823
950395 287044
-351577 -843823
-198755 138512
-306560 -483261
-487474 -857400
885637 -240518
-297576 603522
-748283 33...

output:

3999
7998
11997
15996
19995
23994
27993
31992
35991
39990
43989
47988
51987
55986
59985
63984
67983
71982
75981
79980
83979
87978
91977
95976
99975
103974
107973
111972
115971
119970
123969
127968
131967
135966
139965
143964
147963
151962
155961
159960
163959
167958
171957
175956
179955
183954
18795...

result: