题目大意:求从(0,5)到(10,5)的最短距离,起点与终点之间有n堵墙,每个墙有2个门. 题目思路:判断两点间是否有墙(判断两点的连线是否与某一堵墙的线段相交),建立一个图,然后最短路求出就可以了. #include<cstdio> #include<cstdlib> #include<cmath> #include<iostream> #include<algorithm> #include<cstring> #include&l…

The Doors Time Limit: 1000MS Memory Limit: 10000K Description You are to find the length of the shortest path through a chamber containing obstructing walls. The chamber will always have sides at x = 0, x = 10, y = 0, and y = 10. The initial and fina…

题目链接:https://vjudge.net/problem/POJ-1556 题意:在一个矩形内,起点(0,5)和终点(10,5)是固定的,中间有n个道墙(n<=18),每道墙有两个門,求起点到终点的最短路. 思路: 最多有4*n+2个点,枚举所有点对(p1,p2),用叉积判断线段p1p2和中间的墙是否相交,不相交那么更新距离为两点的距离,否则为inf.更新所有的边之后用floyd得到最短路.答案即dist[0][4*n+1].时间复杂度O(n^3). AC code: #include<…

思路:暴力判断每个点连成的线段是否被墙挡住,构建图.求最短路. 思路很简单,但是实现比较复杂,模版一定要可靠. #include<stdio.h> #include<string.h> #include<math.h> #include<iostream> using namespace std; ,M=N*N; const double INF=0x3f3f3f3f; ; int sgn(double x){ ; ) ; ; } struct point{…

这题就是,处理出没两个点.假设能够到达,就连一条边,推断可不能够到达,利用线段相交去推断就可以.最后求个最短路就可以 代码: #include <cstdio> #include <cstring> #include <algorithm> #include <vector> #include <queue> using namespace std; #include <cstdio> #include <cstring>…

The Doors Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 8334   Accepted: 3218 Description You are to find the length of the shortest path through a chamber containing obstructing walls. The chamber will always have sides at x = 0, x =…

The Doors http://poj.org/problem?id=1556 Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 10466   Accepted: 3891 Description You are to find the length of the shortest path through a chamber containing obstructing walls. The chamber will…

转自:http://blog.csdn.net/shahdza/article/details/7779273 最短路 [HDU] 1548 A strange lift基础最短路(或bfs)★2544 最短路 基础最短路★3790 最短路径问题基础最短路★2066 一个人的旅行基础最短路(多源多汇,可以建立超级源点和终点)★2112 HDU Today基础最短路★1874 畅通工程续基础最短路★1217 Arbitrage 货币交换 Floyd (或者 Bellman-Ford 判环)★124…

出处:http://blog.csdn.net/shahdza/article/details/7779273 最短路 [HDU] 1548    A strange lift基础最短路(或bfs)★ 2544    最短路  基础最短路★ 3790    最短路径问题基础最短路★ 2066    一个人的旅行基础最短路(多源多汇,可以建立超级源点和终点)★ 2112    HDU Today基础最短路★ 1874    畅通工程续基础最短路★ 1217    Arbitrage   货币交换…

1.poj2318 TOYS 传送:http://poj.org/problem?id=2318 题意:有m个点落在n+1个区域内.问落在每个区域的个数. 分析:二分查找落在哪个区域内.叉积判断点与线段的位置. #include<iostream> #include<cstring> #include<algorithm> using namespace std; ; struct point{ int x,y; point(){ } point(int _x,int…