UVa 1643 Angle and Squares (计算几何)

Posted 2020-08-11 dwtfukgv

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;
}