hdu 2988 Dark roads】的更多相关文章

题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=2988 Dark roads Description Economic times these days are tough, even in Byteland. To reduce the operating costs, the government of Byteland has decided to optimize the road lighting. Till now every road…
Dark roads Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1067    Accepted Submission(s): 474 Problem Description Economic times these days are tough, even in Byteland. To reduce the operating…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2988 解题报告:一个裸的最小生成树,没看题,只知道结果是用所有道路的总长度减去最小生成树的长度和. #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; ; ]; struct node { int front,rear,len…
最小生成树: 中文名 最小生成树 外文名 Minimum Spanning Tree,MST 一个有 n 个结点的连通图的生成树是原图的极小连通子图,且包含原图中的所有 n 个结点,并且有保持图连通的最少的边. 最小生成树可以用kruskal(克鲁斯卡尔)算法或prim(普里姆)算法求出. 最小生成树其实是最小权重生成树的简称.   应用: 生成树和最小生成树有许多重要的应用.   例如:要在n个城市之间铺设光缆,主要目标是要使这 n 个城市的任意两个之间都可以通信,但铺设光缆的费用很高,且各个…
最小生成树. /* */ #include <iostream> #include <string> #include <map> #include <queue> #include <set> #include <stack> #include <vector> #include <deque> #include <algorithm> #include <cstdio> #inclu…
Dark roads Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other) Total Submission(s) : 7   Accepted Submission(s) : 2 Problem Description Economic times these days are tough, even in Byteland. To reduce the operating costs,…
题目链接:HDU 1102 Constructing Roads Constructing Roads Problem Description There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are conne…
HDOJ(HDU).1025 Constructing Roads In JGShining's Kingdom (DP) 点我挑战题目 题目分析 题目大意就是给出两两配对的poor city和rich city,求解最多能修几条不相交的路.此题可以转化为LIS问题.转化过程如下: 数据中有2列,为方便表述,暂且叫做第一列和第二列. 1.若第一列是是递增的(给出的2个样例都是递增的),那么要想尽可能多的做连线,则那么就需要找出第二列中最长的递增子序列,若出现非递增的序列,那么连线后一定会相交.…
Dark roads Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1031    Accepted Submission(s): 450 Problem Description Economic times these days are tough, even in Byteland. To reduce the operating…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1025 解题报告:先把输入按照r从小到大的顺序排个序,然后就转化成了求p的最长上升子序列问题了,当然按p排序也是一样的.但这题的n的范围是5*10^5次方,所以用n^2算法求 最长上升子序列肯定不行,下面简单介绍一下nlogn时间内求的方法: 从序列里面每次插入一个数,插入到另外一个数组里面,这个数组初始状态是空的,插入一个数时,如果这个数比这个数组里面的任何一个数都大,则直接插入到最后面,否则判断这…