题目链接 Intersection Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 12040   Accepted: 3125 Description You are to write a program that has to decide whether a given line segment intersects a given rectangle. An example: line: start point:…

题目: Description You are to write a program that has to decide whether a given line segment intersects a given rectangle. An example: line: start point: (4,9) end point: (11,2) rectangle: left-top: (1,5) right-bottom: (7,1)  Figure 1: Line segment doe…

题目链接:http://poj.org/problem?id=1410 Intersection Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 12822   Accepted: 3347 Description You are to write a program that has to decide whether a given line segment intersects a given rectangle.…

Intersection Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 16322   Accepted: 4213 Description You are to write a program that has to decide whether a given line segment intersects a given rectangle. An example: line: start point: (4,9)…

Intersection 大意:给你一条线段,给你一个矩形,问是否相交. 相交:线段全然在矩形内部算相交:线段与矩形随意一条边不规范相交算相交. 思路:知道详细的相交规则之后题事实上是不难的,可是还有个坑点就是题目里明明说给的是矩形左上角跟右下角的点,但实际上不是,须要又一次推断一下...真坑. struct Point { double x, y; } A, B, C, D; struct Line { Point a, b; } L; int n; double xmult(Point p1…

// 线段和矩形相交 POJ 1410 // #include <bits/stdc++.h> #include <iostream> #include <cstdio> #include <cstdlib> #include <algorithm> #include <vector> #include <math.h> using namespace std; #define LL long long typedef p…

题目链接 题意 判断线段和矩形是否有交点(矩形的范围是四条边及内部). 思路 判断线段和矩形的四条边有无交点 && 线段是否在矩形内. 注意第二个条件. Code #include <cstdio> #include <cmath> #include <iostream> #include <cstring> #define inf 0x3f3f3f3f #define eps 1e-6 #define maxn 110 using name…

// 判断线段和直线相交 POJ 3304 // 思路: // 如果存在一条直线和所有线段相交,那么平移该直线一定可以经过线段上任意两个点,并且和所有线段相交. #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include <map> #include <set> #include <queue> #includ…

题目链接:https://vjudge.net/problem/POJ-1410 题意:判断线段和矩形是否相交. 思路:注意这里的相交包括线段在矩形内,因此先判断线段与矩形的边是否相交,再判断线段的两端点是否在矩形内(因为是矩形,即凸多边形,直接用叉积判断即可,如果是一般的多边形,需要用射线法判断.) AC code: #include<cstdio> #include<algorithm> #include<cmath> #include<cstdlib>…

http://poj.org/problem?id=1410 题目大意:给你一个线段和矩形的对角两点  如果相交就输出'T'  不想交就是'F' 注意: 1,给的矩形有可能不是左上 右下  所以要先判断的 2,线段在矩形的内部输出T 3,如果交点是矩形的顶点的话 是不相交的 情况有点多  刚开始考虑的太少   wa的我心疼 #include<stdio.h> #include<string.h> #include<stdlib.h> #include<ctype.…