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//集合中元素是不会重复的,所以完全没有必要将两个集合合并后再进行排序,交换排序的时间效率是O(n^2),将两个集合中的元素分别排序后输出即可.输出格式也非常需要 //注意的.输出一列元素赢以cout<<a[0];然后在循环中输出printf(" %d",a[i]);这样的细节是非常需要加以留心的. #include<iostream> using namespace std; int *swap(int *p,int t); void main() { int…
题目链接:http://lightoj.com/volume_showproblem.php?problem=1412 思路:好久没写题解了,有点手生,这题从昨天晚上wa到现在终于是过了...思想其实很简单,就是预处理出每一块的最长直径,然后每次询问的时候直接查询就可以了. #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<vector&…
网络流/最小割 一开始我是将羊的区域看作连通块,狼的区域看作另一种连通块,S向每个羊连通块连一条无穷边,每个狼连通块向T连一条无穷边,连通块内部互相都是无穷边.其余是四连通的流量为1的边……然后WA了= =自己的数据和样例都过了…… 然后orz了一下Hzwer,改成对每个羊/狼都单独连一条无穷边,分界线/0点周围 连容量1的边……AC…… /************************************************************** Problem: 1412 U…
A Puzzling Problem The goal of this problem is to write a program which will take from 1 to 5 puzzle pieces such as those shown below and arrange them, if possible, to form a square. An example set of pieces is shown here. The pieces cannot be rotate…
 A Puzzling Problem  The goal of this problem is to write a program which will take from 1 to 5 puzzle pieces such as those shown below and arrange them, if possible, to form a square. An example set of pieces is shown here. The pieces cannot be rota…
[题目链接] http://www.lydsy.com/JudgeOnline/problem.php?id=1412 [题意] 在一个n*m的格子中,将羊和狼隔开的最小代价. [思路] 最小割. 由S向狼连边inf,由羊向T连边inf,由狼向空格和羊连边1,由空格向四周连边为1,跑一遍最小割. [代码] #include<cmath> #include<queue> #include<vector> #include<cstdio> #include<…
传送门:http://acm.tzc.edu.cn/acmhome/problemdetail.do?&method=showdetail&id=3151 时间限制(普通/Java):1000MS/3000MS     内存限制:65536KByte 描述 H1N1 like to solve acm problems.But they are very busy, one day they meet a problem. Given three intergers a,b,c, the…
原题链接:http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1412 题目要求判断是否有一条直线可以穿过所有的圆. 做法:把所有圆心做一次凸包,然后判断这个凸包是否能通过一个宽度为2*R的通道. 做法和求凸包直径差不多,只是判断的时候把点到两个端点的距离换成点到直线的距离. #include <stdio.h> #include <string.h> #include <math.h> #include <stdli…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1412 {A} + {B} Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 14595    Accepted Submission(s): 6095 Problem Description 给你两个集合,要求{A} + {B}.注:同一个集…
http://www.lydsy.com/JudgeOnline/problem.php?id=1412 超级源点连向所有的狼,超级汇点连向所有羊,流量为INF 相邻连边流量为1,最小割 #include<cstdio> #include<queue> #include<cstring> #include<algorithm> std::queue<int>que; ]={,,,-,}; #define INF 0x7fffffff ; ][];…
求有限集传递闭包的 Floyd Warshall 算法(矩阵实现) 其实就三重循环.zzuoj 1199 题 链接 http://acm.zzu.edu.cn:8000/problem.php?id=1199 Problem B: 大小关系 Time Limit: 2 Sec  Memory Limit: 128 MBSubmit: 148  Solved: 31[Submit][Status][Web Board] Description 当我们知道一组大小关系之后,可判断所有关系是否都能成立…
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题目链接:http://poj.org/problem?id=3311 题意:一个人到一些地方送披萨,要求找到一条路径能够遍历每一个城市后返回出发点,并且路径距离最短.最后输出最短距离即可.注意:每一个地方可重复访问多次. 经典的状压dp,因为每次送外卖不超过10个地方,可以压缩. 由于题中明确说了两个城市间的直接可达路径(即不经过其它城市结点)不一定是最短路径,所以需要借助floyd首先求出任意两个城市间的最短距离. 然后,在此基础上来求出遍历各个城市后回到出发点的最短路径的距离,即求解TSP…
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