poj 3743 LL’s cake (PSLG，Accepted)
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
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- POJ 3159 Candies（差分约束，最短路）
Candies Time Limit: 1500MS Memory Limit: 131072K Total Submissions: 20067 Accepted: 5293 Descrip ...
- 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 ...
- 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 ...
- [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 ...
- 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
- python实例 条件和循环语句
#! /usr/bin/python #条件和循环语句 x=int(input("Please enter an integer:")) if x<0: x=0 ...
- TP5.1 首页路由
把自带的return 删了
- BootStrap 栅格化换行问题
bootstrap 3的栅格系统有时会出现错位情况,原因是上一行的元素高度不一,导致元素float错误. 解决方法 使用栅格系统的时候,应该在每一行都使用row单独包裹起来,不要直接在一行中连续加入c ...
- BZOJ 3296: [USACO2011 Open] Learning Languages
Time Limit: 5 Sec Memory Limit: 128 MB Submit: 387 Solved: 206 [Submit][Status][Discuss] Description ...
- eclipse修改中文注释的字体（亲测有用！）
Window –> Preferences –> General –> Appearance –> Colors and Fonts –> Basic –> Tex ...
- Katalon系列十八：用例变量&用例间调用
一.用例变量写用例时,我们可以用代码定义变量,如:String name = '新闻'println(name) 上面是硬编码,我们也可以在用例里定义变量,只在该用例里生效哦,想跨用例就用全局变量. ...
- 2018-8-10-win10-UWP-修改密码框文字水平
title author date CreateTime categories win10 UWP 修改密码框文字水平 lindexi 2018-08-10 19:17:19 +0800 2018-2 ...
- onethink二级导航调用
<ul class="nav-main">//添加tree参数 <think:nav name="nav" tree="true&q ...
- 【风马一族_php】数组函数
原文来自:http://www.cnblogs.com/sows/p/6045699.html (博客园的)风马一族 侵犯版本,后果自负 2016-11-09 15:56:26 数组 函数 php- ...
- 日期格式之——new Date()的用法
获取时间: 1 var myDate = new Date();//获取系统当前时间 获取特定格式的时间: 1 myDate.getYear(); //获取当前年份(2位) 2 myDate.getF ...