Stockbroker Grapevine - poj 1125 (Floyd算法)
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 30454 | Accepted: 16659 |
Description
Unfortunately for you, stockbrokers only trust information coming from their "Trusted sources" This means you have to take into account the structure of their contacts when starting a rumour. It takes a certain amount of time for a specific stockbroker to pass the rumour on to each of his colleagues. Your task will be to write a program that tells you which stockbroker to choose as your starting point for the rumour, as well as the time it will take for the rumour to spread throughout the stockbroker community. This duration is measured as the time needed for the last person to receive the information.
Input
Each person is numbered 1 through to the number of stockbrokers. The time taken to pass the message on will be between 1 and 10 minutes (inclusive), and the number of contacts will range between 0 and one less than the number of stockbrokers. The number of stockbrokers will range from 1 to 100. The input is terminated by a set of stockbrokers containing 0 (zero) people.
Output
It is possible that your program will receive a network of connections that excludes some persons, i.e. some people may be unreachable. If your program detects such a broken network, simply output the message "disjoint". Note that the time taken to pass the message from person A to person B is not necessarily the same as the time taken to pass it from B to A, if such transmission is possible at all.
Sample Input
3
2 2 4 3 5
2 1 2 3 6
2 1 2 2 2
5
3 4 4 2 8 5 3
1 5 8
4 1 6 4 10 2 7 5 2
0
2 2 5 1 5
0
Sample Output
3 2
3 10
题意:
首先,题目可能有多组测试数据,每个测试数据的第一行为经纪人数量N(当N=0时,输入数据结束),然后接下来N行描述第i(1<=i<=N)个经纪人与其他经纪人的关系(教你如何画图)。每行开头数字M为该行对应的经纪人有多少个经纪人朋友(该节点的出度,可以为0),然后紧接着M对整数,每对整数表示成a,b,则表明该经纪人向第a个经纪人传递信息需要b单位时间(即第i号结点到第a号结点的孤长为b),整张图为有向图,即弧Vij 可能不等于弧Vji(数据很明显,这里是废话)。当构图完毕后,求当从该图中某点出发,将“消息”传播到整个经纪人网络的最小时间,输出这个经纪人号和最小时间。最小时间的判定方式为——从这个经纪人(结点)出发,整个经纪人网络中最后一个人接到消息的时。如果有一个或一个以上经纪人无论如何无法收到消息,输出“disjoint”。
#include <iostream>
#include<limits.h>
using namespace std;
int num;
int map[][];
void Floyd(){
for(int i=;i<num;i++){
for(int j=;j<num;j++){
for(int k=;k<num;k++){
if(map[j][i]!=INT_MAX&&map[i][k]!=INT_MAX&&j!=k){
if(map[j][k]>map[j][i]+map[i][k]){
map[j][k]=map[j][i]+map[i][k];
}
}
}
}
}
}
int main() {
cin>>num;
while(num){
for(int i=;i<num;i++){
for(int j=;j<num;j++){
map[i][j]=INT_MAX;
}
}
for(int i=;i<num;i++){
int tmp;
cin>>tmp;
for(int j=;j<tmp;j++){
int stocker,time;
cin>>stocker>>time;
map[i][stocker-]=time;
}
}
Floyd();
int min=INT_MAX,max=,st;
for(int i=;i<num;i++){
max=;
for(int j=;j<num;j++){
if(max<map[i][j]&&i!=j){
max=map[i][j];
}
}
if(min>max){
min=max;
st=i+;
}
}
cout<<st<<" "<<min<<endl;
cin>>num;
}
return ;
}
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