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…

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…

传送门 MPI Maelstrom Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5711   Accepted: 3552 Description BIT has recently taken delivery of their new supercomputer, a 32 processor Apollo Odyssey distributed shared memory machine with a hierar…

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…

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…

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…

MPI Maelstrom BIT has recently taken delivery of their new supercomputer, a 32 processor Apollo Odyssey distributed shared memory machine with a hierarchical communication subsystem. Valentine McKee's research advisor, Jack Swigert, has asked her to…

Description BIT has recently taken delivery of their new supercomputer, a 32 processor Apollo Odyssey distributed shared memory machine with a hierarchical communication subsystem. Valentine McKee's research advisor, Jack Swigert, has asked her to be…

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…

题目: BIT has recently taken delivery of their new supercomputer, a 32 processor Apollo Odyssey distributed shared memory machine with a hierarchical communication subsystem. Valentine McKee's research advisor, Jack Swigert, has asked her to benchmark…

题目大意:求点1到所有点最短路径的最大值 思路:水题,单源最短路,网上解题清一色dijkstra,但是点数小于100显然floyd更简洁嘛 #include<cstdio> #include<string.h> #define maxn 101 #define inf 100000 using namespace std; int read(){ int f=1,x=0;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='x…

MPI Maelstrom Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 6499   Accepted: 4036 Description BIT has recently taken delivery of their new supercomputer, a 32 processor Apollo Odyssey distributed shared memory machine with a hierarchic…

MPI Maelstrom Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5547   Accepted: 3458 Description BIT has recently taken delivery of their new supercomputer, a 32 processor Apollo Odyssey distributed shared memory machine with a hierarchic…

题目链接:http://poj.org/problem?id=1502 Description BIT has recently taken delivery of their processor Apollo Odyssey distributed shared memory machine with a hierarchical communication subsystem. Valentine McKee's research advisor, Jack Swigert, has ask…

题目链接:http://poj.org/problem?id=1502 题意:一个处理器给n-1个处理器发送广播,问最短时间.广播时并发,也就是各个路径就大的一方.输入如果是x的话说明两个处理器不能相互通信.输入是矩阵的左三角. 题解:一个最短路的裸题吧.输入的时候注意一下字符的转换.floyd爆一遍之后再对每个路径找最大值即可. 代码: #include<cstdio> #include<cstdlib> #include<cstring> #include<i…

http://poj.org/problem?id=1502 刷一道模板题稳定一下心情... Dijkstra求单源最短路,就是输入的时候注意下,是按下三角输入的(无向图),输入字符x表示i与j不通. 可以这样输入: if(scanf("%d",&w)) map[i][j] = map[j][i] = w; else scanf("x"); #include<stdio.h> #include<string.h> const int…

题意 : 给出 N 个点,各个点之间的路径长度用给出的下三角矩阵表示,上上角矩阵和下三角矩阵是一样的,主对角线的元素都是 0 代表自己到达自己不用花费,现在问你从 1 到 N 的最短路,矩阵的 x 代表点间无法互相到达 分析 : 最短路模板…… 就是在输入的时候需要将字符串变成整数.自己写也可以,也可以使用 atoi(char *)函数,其作用是将字符串数组变成整数,复杂度为 O(n) #include<bits/stdc++.h> using namespace std; ; const i…

题意 给出图,从点1出发,求到最后一个点的时间. 思路 单源最短路,没什么好说的.注意读入的时候的技巧. 代码 #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; const int INF = 1000000000; const int maxn = 110; int n; int edge[maxn][maxn…

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…

MPI Maelstrom 总时间限制:  1000ms 内存限制:  65536kB 描述 BIT has recently taken delivery of their new supercomputer, a 32 processor Apollo Odyssey distributed shared memory machine with a hierarchical communication subsystem. Valentine McKee's research advisor…

POJ.1797 Heavy Transportation (Dijkstra变形) 题意分析 给出n个点,m条边的城市网络,其中 x y d 代表由x到y(或由y到x)的公路所能承受的最大重量为d,求从1到n的所有通路中,所能经过的的最大重量的车为多少. 2. 代码总览 #include <cstdio> #include <cstring> #include <algorithm> #include <queue> #include <stack&…

POJ. 2253 Frogger (Dijkstra ) 题意分析 首先给出n个点的坐标,其中第一个点的坐标为青蛙1的坐标,第二个点的坐标为青蛙2的坐标.给出的n个点,两两双向互通,求出由1到2可行通路的所有步骤当中,步长最大值. 在dij原算法的基础上稍作改动即可.dij求解的是单源最短路,现在求解的是步长最大值,那么更新原则就是,当前的这一步比保存的步如果要大的话,就更新,否则就不更新. 如此求解出来的就是单源最大步骤. 代码总览 #include <cstdio> #include &…

MPI Maelstrom Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 12087   Accepted: 7464 Description BIT has recently taken delivery of their new supercomputer, a 32 processor Apollo Odyssey distributed shared memory machine with a hierarchi…

POJ-1502 MPI Maelstrom 迪杰斯特拉+题解 题意 题意:信息传输,总共有n个传输机,先要从1号传输机向其余n-1个传输机传输数据,传输需要时间,给出一个严格的下三角(其实就是对角线之下的不包括对角线的部分)时间矩阵,a[i][j]代表从i向j传输数据需要的时间,并规定数据传输之间并无影响,即第一个传输机可以同时向其余传输机传输数据.求所有所有的机器都收到消息(他们收到消息后也可以传输)所需的最短时间. 解题思路 这个可以用迪杰斯特拉来求第一台机器到其他所有机器传输消息的时间,…

题目链接:https://vjudge.net/problem/POJ-1502 dijkstra板子题,题目提供下三角情况,不包含正对角线,因为有题意都为0,处理好输入,就是一个很水的题. #include <iostream> #include <cstring> #include <algorithm> #include <cstdio> #include <string> #include <vector> using nam…

#include<cstdio> #include<cstdlib> #include<cstring> #include<iostream> using namespace std; ; const int INF = 0x3f3f3f3f; ]; int n; int f[N][N]; void init() { ; i <= n ; i++) ; j <= n ; j++) if(i == j) f[i][j] = ; else f[i][…

题目链接: http://poj.org/problem?id=2253 Description Freddy Frog is sitting on a stone in the middle of a lake. Suddenly he notices Fiona Frog who is sitting on another stone. He plans to visit her, but since the water is dirty and full of tourists' suns…

(点击此处查看原题) 题目分析 题意:在一个农场中有k台挤奶器和c只奶牛,每个挤奶器最多只能为m只奶牛挤奶,每个挤奶器和奶牛都视为一个点,将编号1~k记为挤奶器的位置,编号k+1~k+c记为奶牛的位置,奶牛只能在这k+c个位置之间移动,输入将给出每个位置和其余k+c个位置的之间道路距离,其中0代表无法到达 问让所有奶牛进行挤奶的情况下(也就是让每头奶牛都走到一个挤奶器的位置上去,而且这个挤奶器上的奶牛不得超过m个),求c只奶牛中走的最远的奶牛的最小移动总距离. 思路:首先思考到这题目要二分答案,…

题目链接:http://poj.org/problem?id=1511 就是求从起点到其他点的最短距离加上其他点到起点的最短距离的和 , 注意路是单向的. 因为点和边很多, 所以用dijkstra优先队列的做法. 起点到其他点的最短距离之和就是dij一下 . 要求其他点到起点的最短距离的话就是把原先边的方向反向一下,然后再求起点到其他点的最短距离之和 , 同样dij一下. 代码如下: #include <iostream> #include <cstdio> #include &l…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=3790 Problem Description 给你n个点,m条无向边,每条边都有长度d和花费p,给你起点s终点t,要求输出起点到终点的最短距离及其花费,如果最短距离有多条路线,则输出花费最少的.   Input 输入n,m,点的编号是1~n,然后是m行,每行4个数 a,b,d,p,表示a和b之间有一条边,且其长度为d,花费为p.最后一行是两个数 s,t;起点s,终点.n和m为0时输入结束.(1<n…