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UVa 1643 Angle and Squares (计算几何)

题意:有n个正方形和一个角(均在第一象限中),使这些正方形与这个角构成封闭的阴影区域,求阴影区域面积的最大值。

析:很容易知道只有所有的正方形的对角形在一条直线时,是最大的,然后根据数学关系,就容易得到答案。

代码如下:

#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cstdio>#include <string>#include <cstdlib>#include <cmath>#include <iostream>#include <cstring>#include <set>#include <queue>#include <algorithm>#include <vector>#include <map>#include <cctype>#include <cmath>#include <stack>#define print(a) printf("%d\n", (a))#define freopenr freopen("in.txt", "r", stdin)#define freopenw freopen("out.txt", "w", stdout)using namespace std;typedef long long LL;typedef pair<int, int> P;const int INF = 0x3f3f3f3f;const double inf = 0x3f3f3f3f3f3f;const LL LNF = 0x3f3f3f3f3f3f;const double PI = acos(-1.0);const double eps = 1e-8;const int maxn = 1e3 + 5;const int mod = 1e9 + 7;const int dr[] = {-1, 0, 1, 0};const int dc[] = {0, 1, 0, -1};const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};int n, m;const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};inline int Min(int a, int b){ return a < b ? a : b; }inline int Max(int a, int b){ return a > b ? a : b; }inline LL Min(LL a, LL b){ return a < b ? a : b; }inline LL Max(LL a, LL b){ return a > b ? a : b; }inline bool is_in(int r, int c){    return r >= 0 && r < n && c >= 0 && c < m;}int main(){    double sq, s, ax, ay, bx, by;    while(scanf("%d", &n) && n){        cin >> ax >> ay >> bx >> by;        sq = 0.0, s = 0.0;        for(int i = 0; i < n; i++){            double m;   cin >> m;            s += m * m;            sq += m;        }        double k1 = ay / ax, k2 = by / bx;        double x1 = fabs((k2+1)*sq/(k2-k1)),  y1 = k1*x1;        double x2 = fabs((k1+1)*sq/(k2-k1)),  y2 = k2*x2;        double area = fabs((x1*y2 - x2*y1) / 2.0);        double sum = area - s/2.0;        printf("%.3lf\n", sum);    }    return 0;}

 

UVa 1643 Angle and Squares (计算几何)