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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
#315152#7730. Convex CheckeryokeffWA 97ms43908kbC++205.3kb2024-01-27 00:31:162024-01-27 00:31:17

Judging History

你现在查看的是最新测评结果

  • [2024-07-04 19:27:17]
  • hack成功,自动添加数据
  • (/hack/727)
  • [2024-07-04 19:17:30]
  • hack成功,自动添加数据
  • (/hack/726)
  • [2024-01-27 00:31:17]
  • 评测
  • 测评结果:WA
  • 用时:97ms
  • 内存:43908kb
  • [2024-01-27 00:31:16]
  • 提交

answer

#include <bits/stdc++.h>

using namespace std;
#define int long long
#define endl '\n'
typedef pair<int, int> PII;
#define fr(i, a, n) for(int i = a; i <= n; i ++)
#define rf(i, a, n) for(int i = n; i >= a; i --)

const int N = 1e6 + 10;
const int mod = 1e9 + 7;
int T, t, n, m, idx, flag, a1, b1, a2, b2, k, x, y, mx, mn;
int mid, mi, c, z, d, e;
int ans, sum, cnt;
int q[N], tr[N * 2], p[N], l[N], r[N], head[N];
string s1, s2, s3, s4, s5;
int ss[30];
vector<string> num[150];
double a, b;

const double eps = 1e-6;

struct point
{
	double x, y;
	point(double x = 0, double y = 0):x(x),y(y){}
};

typedef point Vector;
Vector operator + (Vector a, Vector b)
{
	return Vector(a.x + b.x, a.y + b.y);
}
Vector operator - (Vector a, Vector b)
{
	return Vector(a.x - b.x, a.y - b.y);
}
Vector operator * (Vector a, double p)
{
	return Vector(a.x * p, a.y * p);
}
Vector operator / (Vector a, double p)
{
	return Vector(a.x / p, a.y / p);
}

int sgn(double x)
{
	if(fabs(x) < eps) return 0;
	if(x < 0) return -1;
	else return 1;
}

int dcmp(double x, double y)//点与点比较
{
	if(fabs(x - y) < eps) return 0;
	if(x > y) return 1;
	else return -1;
}

double dot(Vector a, Vector b)//点乘
{
	return a.x * b.x + a.y * b.y;
}

double cross(Vector a, Vector b)//叉乘
{
	return a.x * b.y - a.y * b.x;
}

double length(Vector a)//取模
{
	return sqrt(dot(a, a));
}

double angle(Vector a, Vector b)//夹角,返回弧度制
{
	return acos(dot(a, b) / length(a) / length(b));
}

double area2(point a, point b, point c)//ab与ac构成的有向四边形面积
{
	return cross(b - a, c - a);
}

Vector rotate(Vector a, double rad)//rad为弧度制,且为逆时针旋转的角
{
	return Vector(a.x * cos(rad) - a.y * sin(rad), a.x * sin(rad) + a.y * cos(rad));
}

Vector normal(Vector a)//向量a左转90°的单位法向量
{
	double L = length(a);
	return Vector(-a.y / L, a.x / L);
}

bool tolefttest(point a, point b, point c)//判断bc在ab的逆时针方向(左侧),是返回1,
{
	return cross(b - a, c - b) > 0;
}

point lst[N], qq[N];
int stk[N], top;
bool cmp(point a, point b)
{
	int s = sgn(cross(a - lst[0], b - lst[0]));
	if(s > 0) return true;
	if(s == 0 && length(a - lst[0]) < length(b - lst[0])) return true;
	else return false;
}

bool cmp1(point a, point b)
{
	int s = sgn(cross(a - lst[0], b - lst[0]));
	if(s > 0) return true;
	if(s == 0 && length(a - lst[0]) < length(b - lst[0])) return true;
	else return false;
}
void Graham()
{
	k = 0;
	point p0;
	a = lst[0].x;
	b = lst[0].y;
	for(int i = 1; i < n; i ++)
	{
		if(b > lst[i].y || (b == lst[i].y && a > lst[i].x))
		{
			a = lst[i].x;
			b = lst[i].y;
			k = i;
		}
	}
	if(n == 1)
	{
		top = 1;
		stk[0] = 0;
		return ;
	}
	if(n == 2)
	{
		top = 2, stk[0] = 0, stk[1] = 1;
		return ;
	}
	stk[0] = k, stk[1] = k + 1, top = 2;
//	cout << k << ' ' << top << endl;
	for(int i = k + 2; i < n + k; i ++)
	{
//		cout << i << ' ' << lst[stk[top - 1]].x << ' ' << lst[stk[top - 1]].y << ' ' << lst[stk[top - 2]].x << ' ' << lst[stk[top - 2]].y << ' ' << lst[i].x << " " << lst[i].y << endl;
//		cout << i << ' ' << cross(lst[stk[top - 1]] - lst[stk[top - 2]], lst[i] - lst[stk[top - 2]]) << endl;
		while(top > 1 && cross(lst[stk[top - 1]] - lst[stk[top - 2]], lst[i] - lst[stk[top - 2]]) <= 0)
		{
			top --;
		}
		stk[top] = i;
		top ++;
	}
	if(top == n)
	{
		fr(i, 0, n - 1) qq[i] = lst[i];
		sort(qq, qq + n, cmp);
		flag = 1;
		for(int i = k; i < k + n; i ++)
		{
			if(qq[i - k].x != lst[i].x && lst[i].y != qq[i - k].y) flag = 0;
	 	}
		if(flag == 1)
		{
			top = n;
			return ;
		}
	}
	fr(i, 0, n - 1) qq[i] = lst[n - i - 1];
	fr(i, 0, n - 1) lst[i] = qq[i], lst[i + n] = lst[i];
	k = 0;
	a = lst[0].x;
	b = lst[0].y;
	for(int i = 1; i < n; i ++)
	{
		if(b > lst[i].y || (b == lst[i].y && a > lst[i].x))
		{
			a = lst[i].x;
			b = lst[i].y;
			k = i;
		}
	}
	if(n == 1)
	{
		top = 1;
		stk[0] = 0;
		return ;
	}
	if(n == 2)
	{
		top = 2, stk[0] = 0, stk[1] = 1;
		return ;
	}
	stk[0] = k, stk[1] = k + 1, top = 2;
//	cout << k << ' ' << top << endl;
	for(int i = k + 2; i < n + k; i ++)
	{
//		cout << i << ' ' << lst[stk[top - 1]].x << ' ' << lst[stk[top - 1]].y << ' ' << lst[stk[top - 2]].x << ' ' << lst[stk[top - 2]].y << ' ' << lst[i].x << " " << lst[i].y << endl;
//		cout << i << ' ' << cross(lst[stk[top - 1]] - lst[stk[top - 2]], lst[i] - lst[stk[top - 2]]) << endl;
		while(top > 1 && cross(lst[stk[top - 1]] - lst[stk[top - 2]], lst[i] - lst[stk[top - 2]]) <= 0)
		{
			top --;
		}
		stk[top] = i;
		top ++;
	}
	if(top == n)
	{
		fr(i, 0, n - 1) qq[i] = lst[i];
		sort(qq, qq + n, cmp);
		flag = 1;
		for(int i = k; i < k + n; i ++)
		{
			if(qq[i - k].x != lst[i].x && lst[i].y != qq[i - k].y) flag = 0;
		}
		if(flag == 1)
		{
			top = n;
			return ;
		}
	}
}
void solve()
{
	cin >> n;
	set<PII> st;
	fr(i, 0, n - 1)
	{
		cin >> a >> b;
		st.insert({a, b});
		lst[i] = {a, b};
		lst[i + n] = {a, b};
	} 
	Graham();
//	cout << top << endl;
	if(st.size() >= 3 && top == n)
	{
		cout << "YES" << endl;
	}else cout << "NO" << endl;
}

signed main()
{
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
//	cin >> T;
//	while(T -- )
		solve();
	return 0;
}

詳細信息

Test #1:

score: 100
Accepted
time: 0ms
memory: 34984kb

input:

3
0 0
1 0
0 1

output:

YES

result:

ok answer is YES

Test #2:

score: 0
Accepted
time: 0ms
memory: 35124kb

input:

4
0 0
0 1
1 1
1 0

output:

YES

result:

ok answer is YES

Test #3:

score: 0
Accepted
time: 3ms
memory: 35120kb

input:

4
0 0
0 3
1 2
1 1

output:

YES

result:

ok answer is YES

Test #4:

score: 0
Accepted
time: 4ms
memory: 34964kb

input:

3
0 0
0 0
0 0

output:

NO

result:

ok answer is NO

Test #5:

score: 0
Accepted
time: 0ms
memory: 35136kb

input:

5
1 0
4 1
0 1
2 0
3 2

output:

NO

result:

ok answer is NO

Test #6:

score: 0
Accepted
time: 0ms
memory: 35124kb

input:

5
0 0
1000000000 0
1000000000 500000000
1000000000 1000000000
0 1000000000

output:

NO

result:

ok answer is NO

Test #7:

score: 0
Accepted
time: 3ms
memory: 34948kb

input:

5
0 0
1000000000 0
1000000000 499999999
1000000000 1000000000
0 1000000000

output:

NO

result:

ok answer is NO

Test #8:

score: 0
Accepted
time: 0ms
memory: 35120kb

input:

5
0 0
999999999 0
1000000000 50000000
999999999 1000000000
0 1000000000

output:

YES

result:

ok answer is YES

Test #9:

score: -100
Wrong Answer
time: 97ms
memory: 43908kb

input:

128312
5578014 410408218
5585076 410404717
5588011 410403262
5588473 410403033
5589740 410402405
5593295 410400643
5593751 410400417
5597248 410398684
5598935 410397848
5600618 410397014
5605185 410394751
5610514 410392111
5614281 410390245
5617263 410388768
5621142 410386847
5630840 410382045
56310...

output:

NO

result:

wrong answer expected YES, found NO