ZOJ - 1586 QS Network (Prim) #include<iostream> #include<cstring> using namespace std; +; ;//无穷远 int n; int cost[maxn]; int Edge[maxn][maxn]; int lowcost[maxn]; void Init() { cin>>n; ;i<n;i++) {//读入每个结点的适配器价值 cin>>cost[i]; } ;i&…

题目: In the planet w-503 of galaxy cgb, there is a kind of intelligent creature named QS. QScommunicate with each other via networks. If two QS want to get connected, they need to buy two network adapters (one for each QS) and a segment of network cab…

题目大意: 给出的案例结果得出步骤,如下图所示,从结点1开始查找,找出的一条路径如绿色部分所标注.(关键处在于连接每条路径所需要的适配器的价格得加上去) 代码实现: #include<iostream> #include<cstdio> using namespace std; #define MAX 1000 //注意此处范围得按照题意设置为>=1000,否则会Segmentation Fault #define MAXCOST 0x7fffffff int graph[M…

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=586 题目大意: QS是一种生物,要完成通信,需要设备,每个QS需要的设备的价格不同,并且,这种设备只能在两个QS之间用一次,也就是说,如果一个QS需要和3个QS通信的话,它就必须得买3个设备,同时,对方三个也必须买对应的适合自己的设备.同时,每两个QS之间是有距离的,要完成通信还需要网线,给出每两个QS之间的网线的价值.求一棵生成树,使得所需要的费用最少.数据范围:所有数据都…

QS Network Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu Submit Status Practice ZOJ 1586 Appoint description:  System Crawler  (2015-05-31) Description Sunny Cup 2003 - Preliminary Round April 20th, 12:00 - 17:00 Problem…

QS Network Sunny Cup 2003 - Preliminary Round April 20th, 12:00 - 17:00 Problem E: QS Network In the planet w-503 of galaxy cgb, there is a kind of intelligent creature named QS. QScommunicate with each other via networks. If two QS want to get conne…

最小生成树,刚刚学了Prim算法. 对每条边变的权值进行预处理,c[i][j] = c[i][j] + p[i] + p[j] 其中c[i][j]为输入的权值,p[i],p[j]为连接这两个节点所需的费用. #include<stdio.h> #include<string.h> #include<math.h> #include<algorithm> using namespace std; ; int c[maxn][maxn];//邻接矩阵 int x…

POJ 1502 MPI Maelstrom / UVA 432 MPI Maelstrom / SCU 1068 MPI Maelstrom / UVALive 5398 MPI Maelstrom /ZOJ 1291 MPI Maelstrom (最短路径) Description BIT has recently taken delivery of their new supercomputer, a 32 processor Apollo Odyssey distributed shar…

ZOJ 2477 Magic Cube(魔方) Time Limit: 2 Seconds      Memory Limit: 65536 KB This is a very popular game for children. In this game, there's a cube, which consists of 3 * 3 * 3 small cubes. We can unwrap the cube, it will become like this: 这是个有名的儿童游戏.游戏…

c/c++ 用普利姆(prim)算法构造最小生成树 最小生成树(Minimum Cost Spanning Tree)的概念: ​ 假设要在n个城市之间建立公路,则连通n个城市只需要n-1条线路.这时,自然会考虑,如何在最节省经费的前提下建立这个公路网络. ​ 每2个城市之间都可以设置一条公路,相应地都要付出一定的经济代价.n个城市之间,最多可以设置n(n-1)/2条线路,那么,如何在这些可能的线路中选择n-1条,以使总的耗费最少? 普利姆(prim)算法的大致思路: ​ 大致思想是:设图G顶点…