【CQOI2006】凸多边形

标签：line   ems   --   algorithm   turn   printf   cstring   code   cti   

题面

题解

半平面交模板题

今年\(\mathrm{PKUWC\;D2T3}\)其实只要稍微会一些计算几何的知识就至少有\(76pts\)了

于是准备开始恶补计算几何

其实这道题我从去PKUWC前开始做，做到现在

代码

#include<cstdio>
#include<cstring>
#include<algorithm>
#define RG register
#define clear(x, y) memset(x, y, sizeof(x));

const double INF(1000.);
const int maxn(10010);
struct point { double x, y; };
struct line { point x, y; };
typedef point vector;

inline vector operator + (const vector &lhs, const vector &rhs)
    { return (vector) {lhs.x + rhs.x, lhs.y + rhs.y}; }
inline vector operator - (const vector &lhs, const vector &rhs)
    { return (vector) {lhs.x - rhs.x, lhs.y - rhs.y}; }
inline vector operator * (const vector &lhs, const double &rhs)
    { return (vector) {lhs.x * rhs,   lhs.y * rhs};   }
inline vector operator / (const vector &lhs, const double &rhs)
    { return (vector) {lhs.x / rhs,   lhs.y / rhs};   }
inline double operator * (const vector &lhs, const vector &rhs)
    { return lhs.x * rhs.x + lhs.y * rhs.y; }
inline double cross(const vector &lhs, const vector &rhs)
    { return lhs.x * rhs.y - lhs.y * rhs.x; }

point S[maxn]; int top;
point Intersection(line a, line b)
{
    point c = b.x - a.x;
    double t = cross(b.y, c) / cross(b.y, a.y);
    return a.x + a.y * t;
}

inline void prepare()
{
    top = 0;
    S[++top] = (point) {INF,  INF};
    S[++top] = (point) {-INF, INF};
    S[++top] = (point) {-INF, -INF};
    S[++top] = (point) {INF, -INF};
}

void add_line(line a)
{
    static point tmp[maxn]; int tot = 0;
    S[top + 1] = S[1];
    for(RG int i = 1; i <= top; i++)
    {
        double p = cross(a.y, S[i] - a.x);
        double q = cross(a.y, S[i + 1] - a.x);
        if(p >= 0) tmp[++tot] = S[i];
        if(p * q < 0) tmp[++tot] = Intersection(a, (line) {S[i], S[i + 1] - S[i]});
    }
    top = tot;
    for(RG int i = 1; i <= top; i++) S[i] = tmp[i];
}

point p[maxn];
int main()
{
    int T, n; scanf("%d", &T);
    prepare();
    while(T--)
    {
        scanf("%d", &n);
        for(RG int i = 1; i <= n; i++)
            scanf("%lf%lf", &p[i].x, &p[i].y);
        p[n + 1] = p[1];
        for(RG int i = 1; i <= n; i++)
            add_line((line) {p[i], p[i + 1] - p[i]});
    }
    double ans = 0; S[top + 1] = S[1];
    for(RG int i = 2; i <= top; i++) ans += cross(S[i] - S[1], S[i + 1] - S[1]);
    printf("%.3lf\n", ans / 2.);
    return 0;
}

【CQOI2006】凸多边形

标签：line   ems   --   algorithm   turn   printf   cstring   code   cti   

原文地址：https://www.cnblogs.com/cj-xxz/p/10327208.html