Language: Default Map Labeler Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 1815   Accepted: 599 Description Map generation is a difficult task in cartography. A vital part of such task is automatic labeling of the cities in a map; whe…

题意: 给出n个点  让求这n个点所能建成的正方形的最大边长,要求不覆盖,且这n个点在正方形上或下边的中点位置 解析: 当然是二分,但建图就有点还行..比较难想..行吧...我太垃圾... 2 - sat建图 一看逻辑关系,二看具体情况 这里既有平常常用的逻辑关系,也有一些具体的关系 #include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <…

POJ 2296 Map Labeler / ZOJ 2493 Map Labeler / HIT 2369 Map Labeler / UVAlive 2973 Map Labeler(2-sat 二分) Description Map generation is a difficult task in cartography. A vital part of such task is automatic labeling of the cities in a map; where for e…

POJ 2296 Map Labeler 题目链接 题意: 坐标轴上有N个点.要在每一个点上贴一个正方形,这个正方形的横竖边分别和x,y轴平行,而且要使得点要么在正方形的上面那条边的中点,或者在以下那条边的中点.而且随意两个点的正方形都不重叠(能够重边).问正方形最大边长能够多少? 思路:显然的2-sat问题,注意推断两个矩形相交的地方,细节 代码: #include <cstdio> #include <cstring> #include <cstdlib> #inc…

Map Labeler Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 1267   Accepted: 409 Description Map generation is a difficult task in cartography. A vital part of such task is automatic labeling of the cities in a map; where for each city t…

Map Labeler Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 2047   Accepted: 682 Description Map generation is a difficult task in cartography. A vital part of such task is automatic labeling of the cities in a map; where for each city t…

http://poj.org/problem?id=2296 题意:题意:给你n个点,每个点上都放一个正方形,点只能在正方形的上边或下边的中点上,所有正方形大小一样,不能有面积重叠,求最大的正方形.(n<=100) #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <iostream> using namespace std…

二分答案 + 2-SAT验证,判断正方形是否相交写起来有点烦,思路还是挺简单的. #include<cstdio> #include<cstring> #include<cmath> #include<stack> #include<queue> #include<algorithm> using namespace std; ; struct Point { double x,y; }p[maxn]; int N; stack<…

题目大意: poj2318改个输出 输出 a: b 即有a个玩具的格子有b个 可以先看下poj2318的报告 用map就很方便 #include <cstdio> #include <cmath> #include <string.h> #include <algorithm> #include <map> using namespace std; ; +; double add(double a,double b) { ; return a+b…

题目链接:http://poj.org/problem?id=2296 二分+2sat,每个点的上下两个方向为2sat的两个状态. //STATUS:C++_AC_16MS_536KB #include <functional> #include <algorithm> #include <iostream> //#include <ext/rope> #include <fstream> #include <sstream> #in…