UVA - 1643 Angle and Squares (角度和正方形)(几何)
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题意:第一象限里有一个角,把n(n <= 10)个给定边长的正方形摆在这个角里(角度任意),使得阴影部分面积尽量大。
分析:当n个正方形的对角线在一条直线上时,阴影部分面积最大。
1、通过给定的xa,ya,xb,yb,可求k1,k2。
2、当n个正方形的对角线在一条直线上时,设A(x1,k1*x1),B(x2,k2*x2),
可列方程组:
解得
3、利用叉积算出AOB的面积,再减去正方形面积和的一半。
#pragma comment(linker, "/STACK:102400000, 102400000") #include<cstdio> #include<cstring> #include<cstdlib> #include<cctype> #include<cmath> #include<iostream> #include<sstream> #include<iterator> #include<algorithm> #include<string> #include<vector> #include<set> #include<map> #include<stack> #include<deque> #include<queue> #include<list> #define Min(a, b) ((a < b) ? a : b) #define Max(a, b) ((a < b) ? b : a) const double eps = 1e-8; inline int dcmp(double a, double b) { if(fabs(a - b) < eps) return 0; return a < b ? -1 : 1; } typedef long long LL; typedef unsigned long long ULL; const int INT_INF = 0x3f3f3f3f; const int INT_M_INF = 0x7f7f7f7f; const LL LL_INF = 0x3f3f3f3f3f3f3f3f; const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f; const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1}; const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1}; const int MOD = 1e9 + 7; const double pi = acos(-1.0); const int MAXN = 1000 + 10; const int MAXT = 10000 + 10; using namespace std; struct Point{ double x, y; void set(double xx, double yy){ x = xx; y = yy; } }; double getArea(Point &A, Point &B){ return A.x * B.y - A.y * B.x; } int main(){ int N; while(scanf("%d", &N) == 1){ if(!N) return 0; Point A, B; scanf("%lf%lf%lf%lf", &A.x, &A.y, &B.x, &B.y); double L = 0; double area = 0; for(int i = 0; i < N; ++i){ double l; scanf("%lf", &l); L += l; area += l * l / 2; } double k1 = A.y / A.x; double k2 = B.y / B.x; if(k1 > k2){ swap(k1, k2); } double x1 = (k2 + 1) * L / (k2 - k1); double y1 = k1 * x1; double x2 = (k1 + 1) * L / (k2 - k1); double y2 = k2 * x2; A.set(x1, y1); B.set(x2, y2); double ans = getArea(A, B) / 2 - area; printf("%.3lf\\n", ans); } return 0; }
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