QOJ.ac
QOJ
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 Convex | rxzfn639 | TL | 0ms | 4160kb | C++23 | 3.9kb | 2024-09-18 18:32:50 | 2024-09-18 18:32:51 |
Judging History
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...