[题目链接] http://poj.org/problem?id=1179 [算法] 区间DP [代码] #include <algorithm> #include <bitset> #include <cctype> #include <cerrno> #include <clocale> #include <cmath> #include <complex> #include <cstdio> #inclu…

[题目链接] 点击打开链接 [算法] 线段树扫描线求周长并 [代码] #include <algorithm> #include <bitset> #include <cctype> #include <cerrno> #include <clocale> #include <cmath> #include <complex> #include <cstdio> #include <cstdlib>…

http://poj.org/problem?id=1179 Polygon Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5078   Accepted: 2139 Description Polygon is a game for one player that starts on a polygon with N vertices, like the one in Figure 1, where N=4. Each…

Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5472   Accepted: 2334 Description Polygon is a game for one player that starts on a polygon with N vertices, like the one in Figure 1, where N=4. Each vertex is labelled with an integer and…

Polygon Time Limit: 1000MS   Memory Limit: 10000K Total Submissions:6633   Accepted: 2834 Description Polygon is a game for one player that starts on a polygon with N vertices, like the one in Figure 1, where N=4. Each vertex is labelled with an inte…

POJ2279 Mr. Young's Picture Permutations 题意 Language:Default Mr. Young's Picture Permutations Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 2513 Accepted: 960 Description Mr. Young wishes to take a picture of his class. The students will…

DP中环形处理 对于DP中存在环的情况,大致有两种处理的方法: 对于很多的区间DP来说,很常见的方法就是把原来的环从任意两点断开(注意并不是直接删掉这条边),在复制一条一模一样的链在这条链的后方,当做线性问题来解,即可实现时间复杂度降维. 情况一:将原来的环从任意两点断开,再当做线性问题来解.情况二:添加一些特殊条件,将断开的前后强行连接起来.两种情况取其中的最优解即可.亦可以实现时间复杂度的降维. 本篇博客将对于第一种情况进行分析. 根据一道例题来探讨. POJ 1179 Polygon 题目…

Picture Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3310    Accepted Submission(s): 1723 Problem Description A number of rectangular posters, photographs and other pictures of the same shape…

http://acm.hdu.edu.cn/showproblem.php?pid=1828 Picture Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2135    Accepted Submission(s): 1134 Problem Description A number of rectangular posters,…

[题目] Picture Problem Description A number of rectangular posters, photographs and other pictures of the same shape are pasted on a wall. Their sides are all vertical or horizontal. Each rectangle can be partially or totally covered by the others. The…

Problem Description A number of rectangular posters, photographs and other pictures of the same shape are pasted on a wall. Their sides are all vertical or horizontal. Each rectangle can be partially or totally covered by the others. The length of th…

题目链接 Description A number of rectangular posters, photographs and other pictures of the same shape are pasted on a wall. Their sides are all vertical or horizontal. Each rectangle can be partially or totally covered by the others. The length of the b…

1206 Picture  题目来源: IOI 1998 基准时间限制:2 秒 空间限制:131072 KB 分值: 160 难度:6级算法题  收藏  关注 给出平面上的N个矩形(矩形的边平行于X轴和Y轴),求这些矩形组成的所有多边形的周长之和.     例如:N = 7.(矩形会有重叠的地方).   合并后的多边形:     多边形的周长包括里面未覆盖部分方块的周长. Input 第1行:1个数N.(2 <= N <= 50000) 第2 - N + 1行,每行4个数,中间用空格分隔,分别…

PictureIOI 1998 A number, N (1 <= N < 5000), of rectangular posters, photographs and other pictures of the same shape are pasted on a wall. Their sides are all vertical or horizontal. Each rectangle can be partially or totally covered by the others.…

http://poj.org/problem?id=1175 题目解析: 这个题因为数据的原因可以很水的过,但我因为把1e-8写成了1e-9WA了N遍,一直WA,题目意思很简单就是相似图形(就是求旋转的图形,我不会,但在discuss中看到 可以通过求图形的长度来求)另外常识性问题double数据类型不能直接比较相等,需要借助一个小数比较,那个比较的小数题目一般会提示,例如这题就是1e-8,另外就是 这题的数据是在是忒坑了! #include <iostream> #include <s…

地址:http://poj.org/problem?id=1177 题目: Picture Time Limit: 2000MS   Memory Limit: 10000K Total Submissions: 12905   Accepted: 6817 Description A number of rectangular posters, photographs and other pictures of the same shape are pasted on a wall. Thei…

Picture Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5516    Accepted Submission(s): 2636 Problem Description A number of rectangular posters, photographs and other pictures of the same shape…

Picture Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 12 Accepted Submission(s): 10   Problem Description A number of rectangular posters, photographs and other pictures of the same shape are pa…

学习链接:http://blog.csdn.net/lwt36/article/details/48908031 学习扫描线主要学习的是一种扫描的思想,后期可以求解很多问题. 扫描线求矩形周长并 hdu 1928 Picture Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4795    Accepted Submission(s):…

Picture Time Limit: 2000MS   Memory Limit: 10000K Total Submissions: 12265   Accepted: 6484 Description A number of rectangular posters, photographs and other pictures of the same shape are pasted on a wall. Their sides are all vertical or horizontal…

1845: [Cqoi2005] 三角形面积并 Time Limit: 3 Sec  Memory Limit: 64 MBSubmit: 1664  Solved: 443[Submit][Status][Discuss] Description 给出n个三角形,求它们并的面积. Input 第一行为n(N < = 100), 即三角形的个数 以下n行,每行6个整数x1, y1, x2, y2, x3, y3,代表三角形的顶点坐标.坐标均为不超过10 ^ 6的实数,输入数据保留1位小数 Out…

Picture 题目大意 IOI 1998 求n (<=5000)个矩形 覆盖的图形 的周长(包括洞), 坐标范围[-10000,10000] 题解 一眼离散化+2维线段树,但仔细一想 空间不太够,时间勉强接受 然后目测可能1维线段树+扫描线了? 然后 竟然 裸的扫描线可以过,如下面代码 总数量级上来讲,输入O(n),排序O(n log n),扫描过程O(sum(len周长)) 约5000*20000*4的上限[ 不过USACO给过了, 所以还是线段树好? 从实现来讲,把矩形拆分成x和y方向,靠…

1206 Picture 题目来源: IOI 1998 基准时间限制:2 秒 空间限制:131072 KB 分值: 160 难度:6级算法题  收藏  关注 给出平面上的N个矩形(矩形的边平行于X轴和Y轴),求这些矩形组成的所有多边形的周长之和. 例如:N = 7.(矩形会有重叠的地方). 合并后的多边形: 多边形的周长包括里面未覆盖部分方块的周长. Input 第1行:1个数N.(2 <= N <= 50000) 第2 - N + 1行,每行4个数,中间用空格分隔,分别表示矩形左下和右上端点…

很容易想到枚举第一步切掉的边,然后再计算能够产生的最大值. 联想到区间DP,令dp[i][l][r]为第一步切掉第i条边后从第i个顶点起区间[l,r]能够生成的最大值是多少. 但是状态不好转移,因为操作的符号不仅有‘+’,还有‘*’,加法的话,父区间的最大值显然可以从子区间的最大值相加得出. 乘法的话,父区间的最大值除了由子区间的最大值相乘得出,还可以由子区间的最小值相乘得出. 所以,多定义一维状态. dp[i][l][r][flag]表示第一步切掉第i条边后从第i个顶点起区间[l,r]能够生成…

Comparison of the different algorithms for  Polygon Boolean operations. Michael Leonov 1998 http://www.angusj.com/delphi/clipper.php#screenshots http://www.complex-a5.ru/polyboolean/comp.html http://www.angusj.com/delphi/clipper.php#screenshots Intro…

Point-In-Polygon Algorithm — Determining Whether A Point Is Inside A Complex Polygon © 1998,2006,2007 Darel Rex Finley.  This complete article, unmodified, may be freely distributed for educational purposes. Visit the new page which adds spline curve…

Given a list of points that form a polygon when joined sequentially, find if this polygon is convex (Convex polygon definition). Note: There are at least 3 and at most 10,000 points. Coordinates are in the range -10,000 to 10,000. You may assume the…

首先在利用 GEOGRAPHY::STPolyFromText(@GeoStr, 4326) 这样的函数把字符串转换为Geography类型时,字符串里经纬度的顺序是 “经度[空格]纬度”,即“longitude latitude”. 另外就是从谷歌地图里得到的多边形(polygon)的顶点定义的顺序和Sql Server里Geography类型中的顶点定义顺序是相反的,即一个是顺时针定义,一个是逆时针定义(至于哪个是顺时针,哪个是逆时针,没有细究),所以把这些顶点存到数据库的时候,需要先反转一…

原文: http://tutorials.jenkov.com/svg/polygon-element.html Polyline 虽然说这个 元素我没用过,但是还是蛮强大的,也翻译下 示例 <svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink"> <polyline points="0,0 30,0 15,30" st…

[OpenGL][SharpGL]用Polygon Offset解决z-fighting和stitching问题 本文参考了(http://www.zeuscmd.com/tutorials/opengl/15-PolygonOffset.php),用SharpGL重写了示例代码,您可以点击文末的链接下载. 什么是stitching和z-fighting 在OpenGL中,如果想绘制一个多边形同时绘制其边界,可是先使用多边形模式GL_FILL绘制物体,然后使用多边形模式GL_LINE和不同的颜色…