题意:判断两个多边形是否有面积大于0的公共部分

思路:扫描线基础。

#pragma comment(linker, "/STACK:10240000")
#include <bits/stdc++.h>
using namespace std; #define X first
#define Y second
#define pb push_back
#define mp make_pair
#define all(a) (a).begin(), (a).end()
#define fillchar(a, x) memset(a, x, sizeof(a)) typedef long long ll;
typedef pair<int, int> pii; namespace Debug {
void print(){cout<<endl;}template<typename T>
void print(const T t){cout<<t<<endl;}template<typename F,typename...R>
void print(const F f,const R...r){cout<<f<<" ";print(r...);}template<typename T>
void print(T*p, T*q){int d=p<q?:-;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;}
}
template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);}
template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);}
/* -------------------------------------------------------------------------------- */ const double eps = 1e-10;/** 设置比较精度 **/
struct Real {
double x;
double get() { return x; }
Real(const double &x) { this->x = x; }
Real() {} Real operator + (const Real &that) const { return Real(x + that.x);}
Real operator - (const Real &that) const { return Real(x - that.x);}
Real operator * (const Real &that) const { return Real(x * that.x);}
Real operator / (const Real &that) const { return Real(x / that.x);} Real operator += (const Real &that) { return Real(x += that.x); }
Real operator -= (const Real &that) { return Real(x -= that.x); }
Real operator *= (const Real &that) { return Real(x *= that.x); }
Real operator /= (const Real &that) { return Real(x /= that.x); } bool operator < (const Real &that) const { return x - that.x <= -eps; }
bool operator > (const Real &that) const { return x - that.x >= eps; }
bool operator == (const Real &that) const { return x - that.x > -eps && x - that.x < eps; }
bool operator <= (const Real &that) const { return x - that.x < eps; }
bool operator >= (const Real &that) const { return x - that.x > -eps; }
}; struct Point {
Real x, y;
int read() { return scanf("%lf%lf", &x.x, &y.x); }
Point(const Real &x, const Real &y) { this->x = x; this->y = y; }
Point() {}
Point operator + (const Point &that) const { return Point(this->x + that.x, this->y + that.y); }
Point operator - (const Point &that) const { return Point(this->x - that.x, this->y - that.y); }
Real operator * (const Point &that) const { return x * that.x + y * that.y; }
Point operator * (const Real &that) const { return Point(x * that, y * that); }
Point operator += (const Point &that) { return Point(this->x += that.x, this->y += that.y); }
Point operator -= (const Point &that) { return Point(this->x -= that.x, this->y -= that.y); }
Point operator *= (const Real &that) { return Point(x *= that, y *= that); }
Real cross(const Point &that) const { return x * that.y - y * that.x; }
};
typedef Point Vector; struct Segment {
Point a, b;
Segment(const Point &a, const Point &b) { this->a = a; this->b = b; }
Segment() {}
bool intersect(const Segment &that) const {
Point c = that.a, d = that.b;
Vector ab = b - a, cd = d - c, ac = c - a, ad = d - a, ca = a - c, cb = b - c;
return ab.cross(ac) * ab.cross(ad) < && cd.cross(ca) * cd.cross(cb) < ;
}
Point getLineIntersection(const Segment &that) const {
Vector u = a - that.a, v = b - a, w = that.b - that.a;
Real t = w.cross(u) / v.cross(w);
return a + v * t;
}
}; Point p1[], p2[];
Segment side1[], side2[]; bool cmp(const pair<Segment, int> &a, const pair<Segment, int> &b) {
return a.X.a.x + a.X.b.x < b.X.a.x + b.X.b.x;
} int main() {
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
#endif // ONLINE_JUDGE
int n, m, cas = ;
while (cin >> n) {
for (int i = ; i < n; i ++) {
p1[i].read();
if (i) side1[i - ] = Segment(p1[i - ], p1[i]);
}
side1[n - ] = Segment(p1[n - ], p1[]);
cin >> m;
for (int i = ; i < m; i ++) {
p2[i].read();
if (i) side2[i - ] = Segment(p2[i - ], p2[i]);
}
side2[m - ] = Segment(p2[m - ], p2[]);
/** 得到所有的扫描线并排序去重 **/
vector<Real> Y;
for (int i = ; i < n; i ++) Y.pb(p1[i].y);
for (int i = ; i < m; i ++) Y.pb(p2[i].y);
for (int i = ; i < n; i ++) {
for (int j = ; j < m; j ++) {
if (side1[i].intersect(side2[j])) {
Y.pb(side1[i].getLineIntersection(side2[j]).y);
}
}
}
sort(all(Y));
Y.resize(unique(all(Y)) - Y.begin());
//Debug::print("Y.size=", Y.size());
Real area = ;
for (int i = ; i < Y.size(); i ++) {
vector<pair<Segment, int> > V;
/** 得到扫描线之间的所有线段 **/
for (int j = ; j < n; j ++) {
Real miny = side1[j].a.y, maxy = side1[j].b.y;
if (miny > maxy) swap(miny, maxy);
if (miny <= Y[i - ] && maxy >= Y[i]) {
Point dot1 = side1[j].getLineIntersection(Segment(Point(, Y[i - ]), Point(, Y[i - ])));
Point dot2 = side1[j].getLineIntersection(Segment(Point(, Y[i]), Point(, Y[i])));
V.pb(mp(Segment(dot1, dot2), ));
}
}
for (int j = ; j < m; j ++) {
Real miny = side2[j].a.y, maxy = side2[j].b.y;
if (miny > maxy) swap(miny, maxy);
if (miny <= Y[i - ] && maxy >= Y[i]) {
Point dot1 = side2[j].getLineIntersection(Segment(Point(, Y[i - ]), Point(, Y[i - ])));
Point dot2 = side2[j].getLineIntersection(Segment(Point(, Y[i]), Point(, Y[i])));
V.pb(mp(Segment(dot1, dot2), ));
}
}
sort(all(V), cmp);
//Debug::print("V.size=", V.size());
/** 从左至右统计 **/
bool in1 = , in2 = ;/** 当前延伸的区域是否在多边形内部 **/
for (int i = ; i < V.size(); i ++) {
if (in1 && in2) area += V[i].X.a.x - V[i - ].X.a.x + V[i].X.b.x - V[i - ].X.b.x;
if (V[i].Y) in2 ^= ;
else in1 ^= ;
}
}
printf("Case %d: %s\n", ++ cas, area > ? "Yes" : "No");
}
return ;
}

[UVA Live 12931 Common Area]扫描线的更多相关文章

  1. UVA 10405 Longest Common Subsequence

    题目链接:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=16&p ...

  2. UVA 10405 Longest Common Subsequence (dp + LCS)

    Problem C: Longest Common Subsequence Sequence 1: Sequence 2: Given two sequences of characters, pri ...

  3. UVA 10405 Longest Common Subsequence --经典DP

    最长公共子序列,经典问题.算是我的DP开场题吧. dp[i][j]表示到s1的i位置,s2的j位置为止,前面最长公共子序列的长度. 状态转移: dp[i][j] = 0                 ...

  4. [UVa OJ] Longest Common Subsequence

    This is the classic LCS problem. Since it only requires you to print the maximum length, the code ca ...

  5. UVA 10522 Height to Area(知三角形三高求面积)

    思路:海伦公式, AC代码: #include<bits/stdc++.h> using namespace std; int main() { int n; scanf("%d ...

  6. 【uva 12219】Common Subexpression Elimination(图论--树+自定义比较器+映射+递归)

    题意:如题,用表达式树来表示一个表达式,且消除公共的部分,即用编号表示.编号 K 定义为表达式第 K 个出现的字符串. 解法:先构造表达式树,给每棵子树用(string,left_son,right_ ...

  7. hdu---(Tell me the area)(几何/三角形面积以及圆面积的一些知识)

    Tell me the area Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) ...

  8. HDU 1798 Tell me the area

    http://acm.hdu.edu.cn/showproblem.php?pid=1798 Problem Description     There are two circles in the ...

  9. HDU 1798 Tell me the area (数学)

    题目链接 Problem Description     There are two circles in the plane (shown in the below picture), there ...

随机推荐

  1. 如果这篇文章说不清epoll的本质,那就过来掐死我吧!

    转载自:https://www.toutiao.com/i6683264188661367309/ 目录 一.从网卡接收数据说起 二.如何知道接收了数据? 三.进程阻塞为什么不占用cpu资源? 四.内 ...

  2. 【半译】在ASP.NET Core中创建内部使用作用域服务的Quartz.NET宿主服务

    在我的上一篇文章中,我展示了如何使用ASP.NET Core创建Quartz.NET托管服务并使用它来按计划运行后台任务.不幸的是,由于Quartz.NET API的工作方式,在Quartz作业中使用 ...

  3. Chrome 浏览器安装 ChroPath 插件

    1.下载地址 http://www.cnplugins.com/devtool/chropath/download.html 2.安装方法 a.把下载的文件更改后缀名变为压缩包,然后解压到本地:如下图 ...

  4. golang方法详解

    Go 语言 类型方法是一种对类型行为的封装 .Go 语言的方法非常纯粹, 可以看作特殊类型的函数,其显式地将对象实例或指针作为函数的第一个参数,并且参数可以自己指定,而不强制要求一定是 this或se ...

  5. pytorch 矩阵数据增加维度unsqueeze和降低维度squeeze

    增加一个维度 out.unsqueeze(-1) 降低一个维度 out.squeeze(dim=1)

  6. /uesr/local/hadoop/tmp/mapred有锁

    原因:  /usr/local/hadoop/tmp/mapred  有锁 解决:修改改文件的权限 在终端输入: cd /usr/local/hadoop/tmp sudo chmod 777 map ...

  7. 支持向量机SVM知识梳理和在sklearn库中的应用

    SVM发展史 线性SVM=线性分类器+最大间隔 间隔(margin):边界的活动范围.The margin of a linear classifier is defined as the width ...

  8. netcore 下的policy授权自定义返回结果

    目前一直在用policy做权限校验,但是好像组里需要将返回结果统一,之前用的都是直接继承AuthorizationHandler然后调用context.Fail(),但是这样会导致没办法自定义返回结果 ...

  9. [源码分析]从"UDF不应有状态" 切入来剖析Flink SQL代码生成 (修订版)

    [源码分析]从"UDF不应有状态" 切入来剖析Flink SQL代码生成 (修订版) 目录 [源码分析]从"UDF不应有状态" 切入来剖析Flink SQL代码 ...

  10. Python(3)

    使用除法来缩减数字,使用余数法来计算个数. class Solution: def hammingWeight(self, n: int) -> int: count = 0 while Tru ...