题目链接:http://poj.org/problem?id=1125 主要是读懂题意 然后就很简单了 floyd算法的应用 代码: #include<iostream> #include<cstdlib> #include<cstdio> #include<cstring> using namespace std; #define maxn 110 #define INF 10000100 int dis[maxn][maxn]; int n; int m…

Stockbroker Grapevine Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 31264 Accepted: 17106 Description Stockbrokers are known to overreact to rumours. You have been contracted to develop a method of spreading disinformation amongst the st…

Stockbroker Grapevine Time Limit: 1000MS Memory Limit: 10000K Description Stockbrokers are known to overreact to rumours. You have been contracted to develop a method of spreading disinformation amongst the stockbrokers to give your employer the tact…

Stockbroker Grapevine Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 28231   Accepted: 15659 Description Stockbrokers are known to overreact to rumours. You have been contracted to develop a method of spreading disinformation amongst th…

Description Stockbrokers are known to overreact to rumours. You have been contracted to develop a method of spreading disinformation amongst the stockbrokers to give your employer the tactical edge in the stock market. For maximum effect, you have to…

一.Description Stockbrokers are known to overreact to rumours. You have been contracted to develop a method of spreading disinformation amongst the stockbrokers to give your employer the tactical edge in the stock market. For maximum effect, you have…

题目大意 给定一个图,问从某一个顶点出发,到其它顶点的最短路的最大距离最短的情况下,是从哪个顶点出发?须要多久? (假设有人一直没有联络,输出disjoint) 解题思路 Floyd不解释 代码 #include <cstdio> #include <iostream> #include <algorithm> #include <cstring> using namespace std; const int INF = 1000000000; const…

Description Stockbrokers are known to overreact to rumours. You have been contracted to develop a method of spreading disinformation amongst the stockbrokers to give your employer the tactical edge in the stock market. For maximum effect, you have to…

Floyd算法计算每对顶点之间的最短路径的问题 题目中隐含了一个条件是一个人能够同一时候将谣言传递给多个人 题目终于的要求是时间最短.那么就要遍历一遍求出每一个点作为源点时,最长的最短路径长是多少,再求这些值其中最小的是多少,就是题目所求 #include<bits/stdc++.h> using namespace std; int n,x,p,t; int m[120][120],dist[120][120],Max[120]; void floyd(int n,int m[][120],…

题目大意: 输入n 接下来n行 每行输入m 接下来m对a,b 若干个人之间会传播谣言,但每个人传播给其他人的速度都不一样, 问最快的传播路线(即耗时最短的)中最耗时的一个传播环节. 如果其中有人不在这个链中,则输出disjoin,否则输出最快的传播人和该条传播路线中的最慢的一个传播环节花费时间. Sample Input 32 2 4 3 52 1 2 3 62 1 2 2 253 4 4 2 8 5 31 5 84 1 6 4 10 2 7 5 202 2 5 1 50 Sample Outp…