poj2451 半平面交

题目链接:http://poj.org/problem?id=2451

题意：在（0，10000）*（0，10000）的坐标系上，给定n个半平面，求出它们围成的图形的面积

每个半平面由两点（x1，y1）（x2，y2）确定的直线确定，规定半平面为直线的左边，即存在一点（x，y）

使得(x – x1) * (y – y2) – (x – x2) * (y – y1) = (x1 – x) * (y2 – y) – (x2 – x) * (y1 – y)>= 0成立

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;

const double eps = 1e-8;
int n, pn, dq[20005], top, bot;
struct Point {
    double x, y;
} p[20005];

struct Line {
    Point a, b;
    double angle;
    Line& operator= (Line l) {
       a.x = l.a.x; a.y = l.a.y;
       b.x = l.b.x; b.y = l.b.y;
       angle = l.angle;
       return *this;
    }
} l[20005];

int dblcmp(double k) {
    if (fabs(k) < eps) return 0;
    return k > 0 ? 1 : -1;
}

double multi(Point p0, Point p1, Point p2) {
    return (p1.x-p0.x)*(p2.y-p0.y)-(p1.y-p0.y)*(p2.x-p0.x);
}

bool cmp(const Line& l1, const Line& l2) {
    int d = dblcmp(l1.angle-l2.angle);
    if (!d) return dblcmp(multi(l1.a, l2.a, l2.b)) > 0; //极角相同时，将更靠半平面里面的放在前面
    return d < 0;
}

void addLine(Line& l, double x1, double y1, double x2, double y2) {
    l.a.x = x1; l.a.y = y1;
    l.b.x = x2; l.b.y = y2;
    /*atan2(y,x)返回值的取值范围为-PI到PI，实际上就是根据向量（x,y）确定极角，
      若向量在1，2象限，则值大于0；3，4象限，则值小于0
      所以atan2（1，1）与atan（-1，-1）值不同
    */
    l.angle = atan2(y2-y1, x2-x1);
}

//求交点，写得略复杂
 void getIntersect(Line l1, Line l2, Point& p) {
    double A1 = l1.b.y - l1.a.y;
    double B1 = l1.a.x - l1.b.x;
    double C1 = (l1.b.x - l1.a.x) * l1.a.y - (l1.b.y - l1.a.y) * l1.a.x;
    double A2 = l2.b.y - l2.a.y;
    double B2 = l2.a.x - l2.b.x;
    double C2 = (l2.b.x - l2.a.x) * l2.a.y - (l2.b.y - l2.a.y) * l2.a.x;
    p.x = (C2 * B1 - C1 * B2) / (A1 * B2 - A2 * B1);
    p.y = (C1 * A2 - C2 * A1) / (A1 * B2 - A2 * B1);
}

bool judge(Line l0, Line l1, Line l2) {
     Point p;
     getIntersect(l1, l2, p);
     return dblcmp(multi(p, l0.a, l0.b)) < 0;
}

void HalfPlaneIntersect( ){
    int i, j;
    /*排序是在满足所有半平面A*x+B*y+C>0或（<,<=,>=）,
      也就是所有半平面的符号均相同的情况下对极角进行排序，
      此题中题目对半平面的定义就等价于这个条件
 */
    sort(l, l+n, cmp);
    for (i = 0, j = 0; i < n; i++)
        if (dblcmp(l[i].angle-l[j].angle) > 0) //极角相同时，只保留最靠里面的那条
            l[++j] = l[i];
    n = j + 1;
    dq[0] = 0; //双端队列
    dq[1] = 1;
    top = 1;   //顶部和底部
    bot = 0;
    for (i = 2; i < n; i++) {
        //当栈顶的两条直线交点在当前半平面外部时，弹栈
        while (top > bot && judge(l[i], l[dq[top]], l[dq[top-1]])) top--;
        /*由于求的是一个凸多边形，所以当半平面转过接近一圈时，某个半平面满足上一个while的条件后，
          它又会影响到底部的两条直线，当底部的两条直线的交点，在当前的半平面外部时，底部弹栈
        */
        while (top > bot && judge(l[i], l[dq[bot]], l[dq[bot+1]])) bot++;
        dq[++top] = i; //当前半平面入栈
    }
    //当最顶部的两条直线的交点不在最底部的半平面内时，顶部的那个半平面是多余的，顶部弹栈
    while (top > bot && judge(l[dq[bot]], l[dq[top]], l[dq[top-1]])) top--;
    //当最底部的两条直线的交点不在最顶部的半平面内时，底部的那个半平面是多余的，底部弹栈
    while (top > bot && judge(l[dq[top]], l[dq[bot]], l[dq[bot+1]])) bot++;
    dq[++top] = dq[bot]; //将最底部的半平面放到最顶部来，方便下面求顶点
    for (pn = 0, i = bot; i < top; i++, pn++)
        getIntersect(l[dq[i+1]], l[dq[i]], p[pn]);
}

double getArea() { //叉积求面积
    if (pn < 3) return 0;
    double area = 0;
    for (int i = 1; i < pn-1; i++)
        area += multi(p[0], p[i], p[i+1]);
    if (area < 0) area = -area;
    return area/2;
}

int main()
{
    int i;
    double x1, y1, x2, y2;

    while (scanf ("%d", &n) != EOF) {
        //输入半平面，由一条线段确定，半平面为在线段的哪边题目已经给出
        for (i = 0; i < n; i++) {
             scanf ("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
             addLine(l[i], x1, y1, x2, y2);
        }
        //这四个半平面的添加要满足题目给出的半平面的定义
        addLine(l[n++], 0, 0, 10000, 0);
        addLine(l[n++], 10000, 0, 10000, 10000);
        addLine(l[n++], 10000, 10000, 0, 10000);
        addLine(l[n++], 0, 10000, 0, 0);
        HalfPlaneIntersect();
        printf ("%.1lf\n", getArea());
    }
    return 0;
}

觉得上面的代码可能会比较慢，于是又写了个版本，虽然时间复杂度还是O（n * log（n）），但是细节上处理会更优，会有常数优化吧，但是实际运行结果没有太大区别

贴下代码吧

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;

const double eps = 1e-8;
const int maxn = 20005;

int dq[maxn], top, bot, pn, order[maxn], ln;
struct Point {
    double x, y;
} p[maxn];

struct Line {
    Point a, b;
    double angle;
} l[maxn];

int dblcmp(double k) {
    if (fabs(k) < eps) return 0;
    return k > 0 ? 1 : -1;
}

double multi(Point p0, Point p1, Point p2) {
    return (p1.x-p0.x)*(p2.y-p0.y)-(p1.y-p0.y)*(p2.x-p0.x);
}

bool cmp(int u, int v) {
    int d = dblcmp(l[u].angle-l[v].angle);
    if (!d) return dblcmp(multi(l[u].a, l[v].a, l[v].b)) > 0;
    return d < 0;
}

void getIntersect(Line l1, Line l2, Point& p) {
    double dot1,dot2;
    dot1 = multi(l2.a, l1.b, l1.a);
    dot2 = multi(l1.b, l2.b, l1.a);
    p.x = (l2.a.x * dot2 + l2.b.x * dot1) / (dot2 + dot1);
    p.y = (l2.a.y * dot2 + l2.b.y * dot1) / (dot2 + dot1);
}


bool judge(Line l0, Line l1, Line l2) {
    Point p;
    getIntersect(l1, l2, p);
    return dblcmp(multi(p, l0.a, l0.b)) < 0;
}

void addLine(double x1, double y1, double x2, double y2) {
    l[ln].a.x = x1; l[ln].a.y = y1;
    l[ln].b.x = x2; l[ln].b.y = y2;
    l[ln].angle = atan2(y2-y1, x2-x1);
    order[ln] = ln;
    ln++;
}

void halfPlaneIntersection() {
    int i, j;
    sort(order, order+ln, cmp);
    for (i = 1, j = 0; i < ln; i++)
        if (dblcmp(l[order[i]].angle-l[order[j]].angle) > 0)
            order[++j] = order[i];
    ln = j + 1;
    dq[0] = order[0];
    dq[1] = order[1];
    bot = 0;
    top = 1;
    for (i = 2; i < ln; i++) {
        while (bot < top && judge(l[order[i]], l[dq[top-1]], l[dq[top]])) top--;
        while (bot < top && judge(l[order[i]], l[dq[bot+1]], l[dq[bot]])) bot++;
        dq[++top] = order[i];
    }
    while (bot < top && judge(l[dq[bot]], l[dq[top-1]], l[dq[top]])) top--;
    while (bot < top && judge(l[dq[top]], l[dq[bot+1]], l[dq[bot]])) bot++;
    dq[++top] = dq[bot];
    for (pn = 0, i = bot; i < top; i++, pn++)
        getIntersect(l[dq[i+1]], l[dq[i]], p[pn]);
}

double getArea() {
    if (pn < 3) return 0;
    double area = 0;
    for (int i = 1; i < pn-1; i++)
        area += multi(p[0], p[i], p[i+1]);
    return fabs(area)/2;
}

int main()
{
    int  i;
    double x1, y1, x2, y2;

    while (scanf ("%d", &pn) != EOF) {
        for (ln = i = 0; i < pn; i++) {
            scanf ("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
            addLine(x1, y1, x2, y2);
        }
        addLine(0, 0, 10000, 0);
        addLine(10000, 0, 10000, 10000);
        addLine(10000, 10000, 0, 10000);
        addLine(0, 10000, 0, 0);
        halfPlaneIntersection();
        printf ("%.1lf\n", getArea());
    }
    return 0;
}