3743 -- LL’s cake

  搞了好久都过不了,看了下题解是用PSLG来做的。POJ 2164 && LA 3218 Find the Border (Geometry, PSLG 平面直线图) - LyonLys - 博客园 这篇里面写过一下,就是把点都提取出来,然后模拟沿着边界移动,找到多边形并计算面积。

  而我的做法是直接模拟多边形切割,各种超时爆内存。先留着,看以后能不能用这个来过。

没过的代码:

 #include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <queue> using namespace std; const double EPS = 1e-;
inline int sgn(double x) { return (x > EPS) - (x < -EPS);}
struct Point {
double x, y;
Point() {}
Point(double x, double y) : x(x), y(y) {}
Point operator + (Point a) { return Point(x + a.x, y + a.y);}
Point operator - (Point a) { return Point(x - a.x, y - a.y);}
Point operator * (double p) { return Point(x * p, y * p);}
Point operator / (double p) { return Point(x / p, y / p);}
bool operator < (Point a) const { return sgn(x - a.x) < || sgn(x - a.x) == && y < a.y;}
bool operator == (Point a) const { return sgn(x - a.x) == && sgn(y - a.y) == ;}
} ; inline double cross(Point a, Point b) { return a.x * b.y - a.y * b.x;}
inline double dot(Point a, Point b) { return a.x * b.x + a.y * b.y;}
inline double veclen(Point x) { return sqrt(dot(x, x));}
inline Point normal(Point x) { return Point(-x.y, x.x) / veclen(x);}
inline Point vecunit(Point x) { return x / veclen(x);} struct Line {
Point s, t;
Line() {}
Line(Point s, Point t) : s(s), t(t) {}
Point vec() { return t - s;}
Point point(double x) { return s + vec() * x;}
} ;
inline Point llint(Line a, Line b) { return a.point(cross(b.vec(), a.s - b.s) / cross(a.vec(), b.vec()));}
inline bool onseg(Point x, Point s, Point t) { return sgn(cross(s - x, t - x)) == && sgn(dot(s - x, t - x)) < ;}
inline bool onseg(Point x, Line a) { return onseg(x, a.s, a.t);} struct Circle {
Point c;
double r;
Circle() {}
Circle(Point c, double r) : c(c), r(r) {}
bool in(Point x) { return sgn(veclen(x - c) - r) <= ;}
Point point(double x) { return Point(c.x + cos(x) * r, c.y + sin(x) * r);}
} ;
const double R = 10.0;
Circle cake = Circle(Point(0.0, 0.0), R);
const double PI = acos(-1.0);
template<class T> T sqr(T x) { return x * x;}
inline double angle(Point x) { return atan2(x.y, x.x);} int clint(Line s, Point *sol) {
Point nor = normal(s.vec()), ip = llint(s, Line(cake.c, cake.c + nor));
double dis = veclen(cake.c - ip);
if (sgn(dis - cake.r) >= ) return ;
Point dxy = vecunit(s.vec()) * sqrt(sqr(cake.r) - sqr(dis));
int ret = ;
sol[ret] = ip + dxy;
if (onseg(sol[ret], s)) ret++;
sol[ret] = ip - dxy;
if (onseg(sol[ret], s)) ret++;
return ret;
} double getsec(Point a, Point b) {
double a1 = angle(a - cake.c);
double a2 = angle(b - cake.c);
double da = fabs(a1 - a2);
if (da > PI) da = PI * 2.0 - da;
return sqr(cake.r) * da * sgn(cross(a - cake.c, b - cake.c)) / 2.0;
} inline double gettri(Point a, Point b) { return cross(a - cake.c, b - cake.c) / 2.0;}
//typedef vector<Point> VP;
const int N = ;
struct VP {
Point vex[N];
int n;
void clear() { n = ;}
void push_back(Point x) { vex[n++] = x;}
void pop_back() { n--;}
int size() { return n;}
} ; double cpint(VP pt) {
double ret = 0.0;
int n = pt.size();
Point tmp[];
pt.vex[n] = pt.vex[];
for (int i = ; i < n; i++) {
int ic = clint(Line(pt.vex[i], pt.vex[i + ]), tmp);
if (ic == ) {
if (!cake.in(pt.vex[i]) || !cake.in(pt.vex[i + ])) ret += getsec(pt.vex[i], pt.vex[i + ]);
else ret += gettri(pt.vex[i], pt.vex[i + ]);
} else if (ic == ) {
if (cake.in(pt.vex[i])) ret += gettri(pt.vex[i], tmp[]), ret += getsec(tmp[], pt.vex[i + ]);
else ret += getsec(pt.vex[i], tmp[]), ret += gettri(tmp[], pt.vex[i + ]);
} else {
if (pt.vex[i] < pt.vex[i + ] ^ tmp[] < tmp[]) swap(tmp[], tmp[]);
ret += getsec(pt.vex[i], tmp[]);
ret += gettri(tmp[], tmp[]);
ret += getsec(tmp[], pt.vex[i + ]);
}
// cout << "~~ic " << ic << ' ' << ret << endl;
}
return fabs(ret);
} bool fixpoly(VP &poly) {
double sum = 0.0;
int n = poly.size();
poly.vex[n] = poly.vex[];
for (int i = ; i < n; i++) sum += cross(poly.vex[i], poly.vex[i + ]);
if (sgn(sum) == ) return false;
if (sgn(sum) < ) reverse(poly.vex, poly.vex + n);
return true;
} void cutpoly(VP &poly, Line l, VP &ret) {
ret.clear();
int n = poly.size();
// cout << n << endl;
poly.vex[n] = poly.vex[];
for (int i = ; i < n; i++) {
if (sgn(cross(l.vec(), poly.vex[i] - l.s)) >= ) ret.push_back(poly.vex[i]);
if (sgn(cross(l.vec(), poly.vex[i] - poly.vex[i + ]))) {
Point ip = llint(l, Line(poly.vex[i], poly.vex[i + ]));
// cout << "ip " << ip.x << ' ' << ip.y << endl;
if (onseg(ip, poly.vex[i], poly.vex[i + ]) || poly.vex[i] == ip) ret.push_back(ip);
}
}
// cout << "cp sz " << ret.size() << endl;
} const int M = ;
int q[], qh, qt, nu;
VP rec[M];
queue<int> recycle; int getID() {
int ret;
if (nu >= M) {
if (recycle.empty()) { puts("shit!"); while () ;}
ret = recycle.front();
recycle.pop();
} else ret = nu++;
return ret;
} void retID(int x) { recycle.push(x);} int main() {
// freopen("in", "r", stdin);
// freopen("out", "w", stdout);
int T, n, tmp;
double x, y;
cin >> T;
while (T-- && cin >> n) {
while (!recycle.empty()) recycle.pop();
qh = qt = nu = ;
tmp = getID();
rec[tmp].clear();
rec[tmp].push_back(Point(-R * 2.0, -R * 2.0));
rec[tmp].push_back(Point(R * 2.0, -R * 2.0));
rec[tmp].push_back(Point(R * 2.0, R * 2.0));
rec[tmp].push_back(Point(-R * 2.0, R * 2.0));
fixpoly(rec[tmp]);
q[qt++] = tmp;
for (int i = ; i < n; i++) {
cin >> x >> y;
int sz = qt - qh;
Line t = Line(cake.point(x), cake.point(y));
// cout << cake.point(x).x << '=' << cake.point(x).y << endl;
// cout << cake.point(y).x << '~' << cake.point(y).y << endl;
for (int j = ; j < sz; j++) {
tmp = getID();
// cout << "qh ?? " << qh << ' ' << q[qh] << ' ' << rec[q[qh]].size() << endl;
cutpoly(rec[q[qh]], t, rec[tmp]);
if (fixpoly(rec[tmp])) {
// cout << j << "~~1 " << rec[tmp].size() << endl;
// for (int k = 0; k < rec[tmp].size(); k++) cout << rec[tmp].vex[k].x << ' ' << rec[tmp].vex[k].y << endl;
q[qt++] = tmp;
}
swap(t.s, t.t);
tmp = getID();
cutpoly(rec[q[qh]], t, rec[tmp]);
if (fixpoly(rec[tmp])) {
// cout << j << "~~2 " << rec[tmp].size() << endl;
// for (int k = 0; k < rec[tmp].size(); k++) cout << rec[tmp].vex[k].x << ' ' << rec[tmp].vex[k].y << endl;
q[qt++] = tmp;
}
retID(q[qh++]);
}
// cout << "sz~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ " << qt - qh << endl;
}
double mx = 0.0;
while (qh < qt) {
mx = max(mx, cpint(rec[q[qh++]]));
// cout << ".. " << mx << endl;
}
printf("%.2f\n", mx);
}
return ;
} /*
6
2
-3.140000 0.000000
-1.000000 1.000000
2
-3.141592 0.000000
-1.570796 1.570796
3
-3.000000 3.000000
-2.000000 2.000000
-1.000000 1.000000
4
-3.140000 0.000000
-1.000000 1.000000
-3.140000 -1.000000
1.000000 0.000000
6
-3.140000 0.000000
-1.000000 1.000000
-3.140000 -1.000000
1.000000 0.000000
-3.140000 -1.000000
1.000000 0.000000
6
-3.141592 0.000000
-1.570796 1.570796
-3.141592 -1.570796
0.000000 1.570796
-3.141592 1.570796
0.000000 -1.570796
*/

PSLG的方法将尽快更新上来!

UPD:

  模拟遍历边界,985ms压线过,因为有圆弧,所以有几个特判。比较好奇别人那些稳稳的不超时除了少用了STL还做了些什么?

代码如下:

 #include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <vector>
#include <map>
#include <set> using namespace std; const double EPS = 1e-;
inline int sgn(double x) { return (x > EPS) - (x < -EPS);} struct Point {
double x, y;
int id;
Point() {}
Point(double x, double y) : x(x), y(y) {}
bool operator < (Point a) const { return sgn(x - a.x) < || sgn(x - a.x) == && y < a.y;}
bool operator == (Point a) const { return sgn(x - a.x) == && sgn(y - a.y) == ;}
Point operator + (Point a) { return Point(x + a.x, y + a.y);}
Point operator - (Point a) { return Point(x - a.x, y - a.y);}
Point operator * (double p) { return Point(x * p, y * p);}
Point operator / (double p) { return Point(x / p, y / p);}
} ; inline double cross(Point a, Point b) { return a.x * b.y - a.y * b.x;}
inline double dot(Point a, Point b) { return a.x * b.x + a.y * b.y;}
inline double veclen(Point x) { return sqrt(dot(x, x));}
inline Point vecunit(Point x) { return x / veclen(x);}
inline Point normal(Point x) { return Point(-x.y, x.x) / veclen(x);} const int N = ;
Point pts[N * N];
int ptcnt; struct Line {
Point s, t;
Line() {}
Line(Point s, Point t) : s(s), t(t) {}
Point vec() { return t - s;}
Point point(double p) { return s + vec() * p;}
} ; inline Point llint(Line a, Line b) { return a.point(cross(b.vec(), a.s - b.s) / cross(a.vec(), b.vec()));}
inline bool onseg(Point x, Point a, Point b) { return sgn(cross(a - x, b - x)) == && sgn(dot(a - x, b - x)) <= ;}
inline bool onseg(Point x, Line l) { return onseg(x, l.s, l.t);} const double R = 10.0;
inline bool oncircle(Point x) { return sgn(veclen(x) - R) == ;}
inline Point getpt(double p) { return Point(cos(p) * R, sin(p) * R);}
inline double angle(Point x) { return atan2(x.y, x.x);} struct Node {
double ang;
int id;
bool arc;
Node() {}
Node(double ang, int id) : ang(ang), id(id) { arc = false;}
bool operator < (Node x) const { return sgn(ang - x.ang) < || sgn(ang - x.ang) == && arc > x.arc;}
} ;
Line cut[N]; const double PI = acos(-1.0);
template<class T> T sqr(T x) { return x * x;}
Point ori, tmp[N << ];
vector<Node> nb[N * N], oc;
typedef pair<int, int> PII;
typedef pair<int, bool> PIB;
set<PII> used;
map<int, PIB> nx[N * N], anx[N * N]; inline double caltri(Point a, Point b) { return cross(a, b) / 2.0;}
double calsec(Point a, Point b) {
double da = atan2(b.y, b.x) - atan2(a.y, a.x);
da += da < ? PI * 2.0 : 0.0;
return sqr(R) * da / 2.0;
} int main() {
// freopen("in", "r", stdin);
// freopen("out", "w", stdout);
int T, n;
double s, t;
Point ip;
scanf("%d", &T);
while (T-- && ~scanf("%d", &n)) {
ptcnt = ;
used.clear();
for (int i = ; i < n; i++) {
scanf("%lf%lf", &s, &t);
cut[i] = Line(getpt(s), getpt(t));
pts[ptcnt++] = getpt(s);
pts[ptcnt++] = getpt(t);
for (int j = ; j < i; j++) {
if (sgn(cross(cut[i].vec(), cut[j].vec()))) {
ip = llint(cut[i], cut[j]);
// cout << "ip " << ip.x << ' ' << ip.y << endl;
if (onseg(ip, cut[i])) pts[ptcnt++] = ip;
}
}
}
// cout << "pt " << ptcnt << endl;
sort(pts, pts + ptcnt);
ptcnt = unique(pts, pts + ptcnt) - pts;
// cout << "npt " << ptcnt << endl;
for (int i = ; i <= ptcnt; i++) nb[pts[i - ].id = i].clear(), nx[i].clear(), anx[i].clear();
int ptn;
for (int i = ; i < n; i++) {
ptn = ;
for (int j = ; j <= ptcnt; j++) {
if (onseg(pts[j - ], cut[i])) tmp[ptn++] = pts[j - ];
}
sort(tmp, tmp + ptn);
for (int j = ; j < ptn; j++) {
nb[tmp[j].id].push_back(Node(angle(tmp[j - ] - tmp[j]), tmp[j - ].id));
nb[tmp[j - ].id].push_back(Node(angle(tmp[j] - tmp[j - ]), tmp[j].id));
}
}
oc.clear();
for (int i = ; i <= ptcnt; i++) if (oncircle(pts[i - ])) oc.push_back(Node(angle(pts[i - ]), i));
sort(oc.begin(), oc.end());
// for (int i = 0; i < oc.size(); i++) cout << oc[i].id << ' '; cout << endl;
oc.push_back(oc[]);
for (int i = , sz = oc.size(); i < sz; i++) {
nb[oc[i].id].push_back(Node(angle(pts[oc[i - ].id - ] - pts[oc[i].id - ]), oc[i - ].id));
nb[oc[i].id][nb[oc[i].id].size() - ].arc = true;
nb[oc[i - ].id].push_back(Node(angle(pts[oc[i].id - ] - pts[oc[i - ].id - ]), -oc[i].id));
nb[oc[i - ].id][nb[oc[i - ].id].size() - ].arc = true;
}
for (int i = ; i <= ptcnt; i++) {
sort(nb[i].begin(), nb[i].end());
// cout << i << " : " << pts[i - 1].x << ' ' << pts[i - 1].y << endl;
nb[i].push_back(nb[i][]);
// for (int j = 0; j < nb[i].size(); j++) cout << nb[i][j].id << '-' << nb[i][j].arc << ' '; cout << endl;
for (int j = , sz = nb[i].size(); j < sz; j++) {
if (nb[i][j].id < ) continue;
if (nb[i][j].arc) {
if (nb[i][j - ].arc) { if (j < nb[i].size() - ) nx[nb[i][j + ].id][i] = PIB(abs(nb[i][j - ].id), true), j++; else nx[nb[i][].id][i] = PIB(abs(nb[i][j - ].id), true);}
else anx[nb[i][j].id][i] = PIB(abs(nb[i][j - ].id), false);
} else {
if (!nb[i][j - ].arc || nb[i][j - ].id < ) nx[nb[i][j].id][i] = PIB(abs(nb[i][j - ].id), nb[i][j - ].arc);
}
}
nb[i].pop_back();
}
// for (int i = 1; i <= ptcnt; i++) {
// if (anx[i].size()) cout << anx[i].size() << '~' << (*anx[i].begin()).first << '~' << (*anx[i].begin()).second.first << ' ' << nx[i].size() << endl;
// else puts("~~~");
// }
double mx = 0.0, area;
int ls, cur;
PIB tt;
bool arc;
for (int i = ; i < ptcnt; i++) {
for (int j = , sz = nb[i].size(); j < sz; j++) {
if (nb[i][j].arc) continue;
ls = i, cur = nb[i][j].id;
if (used.find(PII(ls, cur)) != used.end()) continue;
arc = false;
area = caltri(pts[ls - ], pts[cur - ]);
used.insert(PII(ls, cur));
// cout << "start " << ls << ' ';
int cnt = ;
while (cur != i && cnt--) {
// cout << cur << ' ';
if (arc) tt = anx[ls][cur];
else tt = nx[ls][cur];
ls = cur, cur = tt.first, arc = tt.second;
if (arc) area += calsec(pts[ls - ], pts[cur - ]);
else area += caltri(pts[ls - ], pts[cur - ]), used.insert(PII(ls, cur));
}
// cout << area << endl;
mx = max(mx, fabs(area));
}
}
printf("%.2f\n", mx);
}
return ;
}

——written by Lyon

poj 3743 LL’s cake (PSLG,Accepted)的更多相关文章

  1. POJ 3743 LL’s cake(圆+PSLG)

    题意是给你一块在原点半径为10的圆,然后告诉你一条直线在圆弧上的极角,相当于用这条直线把这个圆分成两半,然后一共是n条直线切圆,就好比切蛋糕,问你其中最大一块的面积是多少. 如果我们将圆弧转化成直线边 ...

  2. POJ 2828 Buy Tickets(排队问题,线段树应用)

    POJ 2828 Buy Tickets(排队问题,线段树应用) ACM 题目地址:POJ 2828 Buy Tickets 题意:  排队买票时候插队.  给出一些数对,分别代表某个人的想要插入的位 ...

  3. POJ 2135 Farm Tour (网络流,最小费用最大流)

    POJ 2135 Farm Tour (网络流,最小费用最大流) Description When FJ's friends visit him on the farm, he likes to sh ...

  4. POJ 2516 Minimum Cost (网络流,最小费用流)

    POJ 2516 Minimum Cost (网络流,最小费用流) Description Dearboy, a goods victualer, now comes to a big problem ...

  5. POJ 3159 Candies(差分约束,最短路)

    Candies Time Limit: 1500MS   Memory Limit: 131072K Total Submissions: 20067   Accepted: 5293 Descrip ...

  6. POJ 3259 Wormholes(最短路,判断有没有负环回路)

    Wormholes Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 24249   Accepted: 8652 Descri ...

  7. POJ 2175 Evacuation Plan (费用流,负环,消圈法,SPFA)

    http://poj.org/problem?id=2175 Evacuation Plan Time Limit: 1000MS   Memory Limit: 65536K Total Submi ...

  8. [ACM] POJ 3687 Labeling Balls (拓扑排序,反向生成端)

    Labeling Balls Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 10161   Accepted: 2810 D ...

  9. POJ 3154 Graveyard【多解,数论,贪心】

    Graveyard Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 1707   Accepted: 860   Specia ...

随机推荐

  1. 在虚拟机Ubuntu14.04中设置静态ip后无法上网的问题的解决

    背景:用着实验室代理服务器上网. 原因:网关和DNS没有设置正确 网关和DNS参照 真机cmd在命令窗口使用ipconfig  /all命令查看   虚拟机: sudo gedit /etc/netw ...

  2. vuecli3取消eslint

    1.之前好好的项目,今天运行npm run serve.忽然报错运行不了了. 2.原因 在生成项目时候没有忽略 ESlint选项配置,其实建议创建项目的时候还是忽略掉这个选项比较好,因为后期写项目多空 ...

  3. Ajax--serialize应用表单数据序列化

    一.jQuery+Ajax表单数据序列化 <!DOCTYPE html> <html> <head> <meta charset="UTF-8&qu ...

  4. js面向对象开发基础

    js的面向对象开发能力较弱,基本是以prototype为核心的面向对象,虽然现在出了个class这玩意,但本文还是先不做探讨. 面向对象基础——构造函数方法 var Fly = function (s ...

  5. hive-oracle-常用分析函数-窗口函数

    常用的分析函数如下所列: row_number() over(partition by ... order by ...)rank() over(partition by ... order by . ...

  6. Http响应response(文件下载、验证码)

    Http响应response response:响应 作用: 往浏览器写东西 组成部分: 响应行 响应头 响应体 操作响应行 格式: 协议/版本 状态码 状态码说明 状态码: 1xx:已发送请求 2x ...

  7. 关于PHP学习--摘自知乎

    主要是学框架(其实也没啥可学的).数据库.服务器.linux. 所以我推荐apache/nginx文档,框架的文档,mysql的文档,linux使用说明,等等等等. PHP: PHP 手册MySQL ...

  8. 外贸电子商务网站之Prestashop修改顶部导航

    如修改以上所示顶部导航. 如何在prestashop顶部导航栏添加链接,Module>Top horizontal menu点击进入Configure页面 1,在Settings 中看到 链接 ...

  9. node项目搭建

    一:安装 1.简单安装法 下载.msi [编译好的nodejs]  ->  点击安装 [系统会自动配置系统变量]   2.复杂安装法(不推荐) 由于nodejs的源码由C++和js组成 同时需要 ...

  10. 第一次作业:reading and prepare

    这个作业属于哪个课程 课程的链接 这个作业要求在哪里 作业要求的链接 我在这个课程的目标是 理解软件开发流程,更好的开发自己的软件 这个作业在哪个具体方面帮助我实现目标 对"工程" ...