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…

链接:http://poj.org/problem?id=1502 MPI Maelstrom Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5249   Accepted: 3237 Description BIT has recently taken delivery of their new supercomputer, a 32 processor Apollo Odyssey distributed share…

题目链接:http://poj.org/problem?id=1502 题意是给你n个点,然后是以下三角的形式输入i j以及权值,x就不算 #include <iostream> #include <cstdio> #include <cstring> #include <queue> using namespace std; ; const int INF = 1e9; typedef pair <int , int> P; struct da…

MPI Maelstrom Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 4017   Accepted: 2412 Description BIT has recently taken delivery of their new supercomputer, a 32 processor Apollo Odyssey distributed shared memory machine with a hierarchic…

一道不算太难的最短路喵~ 容我吐槽一下,酋长的地位居然不是最高的额——那你特么的居然还算是酋长?! 枚举一个地位区间 [i..i+M-1] 只要所有的交易者的地位都在该区间中,那么就不会引起冲突 而这个可悲的酋长是必须在区间中的,所以若酋长的地位为 L0 那么该枚举的区间就是 [L0-i, L0+M-i] {i|0<=i<=M} 然后就是裸的 dijkstra 了,取出地位在区间中的点,然后新增一个点 N+1,向所有点连边,距离是所有物品的直接价格 若 物品 i 可以通过 物品 j 来降价,那…

MPI Maelstrom Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 20000/10000K (Java/Other) Total Submission(s) : 2   Accepted Submission(s) : 1 Problem Description BIT has recently taken delivery of their new supercomputer, a 32 processor Apollo…

尽管题目很长 写的很玄乎 让我理解了半天 但是事实上就是个模板题啊摔 一发水过不解释 #include <iostream> #include <string> #include <cstdio> #include <cmath> #include <cstring> #include <queue> #include <map> #include <vector> #include <set> #…

MPI Maelstrom Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 6044   Accepted: 3761 Description BIT has recently taken delivery of their new supercomputer, a 32 processor Apollo Odyssey distributed shared memory machine with a hierarchic…

题目大意: 给你 1到n ,  n个计算机进行数据传输, 问从1为起点传输到所有点的最短时间是多少, 其实就是算 1 到所有点的时间中最长的那个点. 然后是数据 给你一个n 代表有n个点, 然后给你一个邻接矩阵, 只有一半,另一半自己补 下面是练习的代码. 分别用了Floyd 和 Dijkstra 还有 Spfa(邻接矩阵版) #include <iostream> #include <cmath> #include <cstring> #include <cst…

#include <cstdio> #include <iostream> #include <stdlib.h> #include <memory.h> using namespace std; ; << ; int s[maxn][maxn],dis[maxn],visit[maxn],n; void dijstra() { memset(visit,,sizeof(visit)); ;i<=n;i++) dis[i]=s[][i];…