poj 1125 Stockbroker Grapevine
Input
Your program will input data for different sets of stockbrokers. Each set starts with a line with the number of stockbrokers. Following this is a line for each stockbroker which contains the number of people who they have contact with, who these people are, and the time taken for them to pass the message to each person. The format of each stockbroker line is as follows: The line starts with the number of contacts (n), followed by n pairs of integers, one pair for each contact. Each pair lists first a number referring to the contact (e.g. a '1' means person number one in the set), followed by the time in minutes taken to pass a message to that person. There are no special punctuation symbols or spacing rules.
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
For each set of data, your program must output a single line containing the person who results in the fastest message transmission, and how long before the last person will receive any given message after you give it to this person, measured in integer minutes.
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.
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it's a basic dijkstra shortest path problem.
dijkstra algorithm :
choose a source vertex and set the its value 0
other vertices are set to infinite
while(Q != empty){
pick u = extract-min(Q)
for each v in adj(u) && v still in Q{
if(d(v) > d(u) + w(u,v)){
d(v)= d(u)+w(u,v)
parent(v) = u //record
}
}
}
although it's not complicate, I still feel unfamiliar with the structure of graph. actually, the data structure I use in this problem is quite simple and not practical at all. however, something in my implementation is still redundant :)......
代码 struct edge { int from; int to; // vertex int edge_value;}; struct broker_info { int num; int path_value; // value of the vetex map < int , edge > adjacency; // all the adjecent vertex from this broker };map < int , broker_info > broker1; // the information of this broker
another deficiency is that I didn't use heap or F tree store the graph，which makes the time complicity a little bit higher.
the STL has a heap template: make_heap() etc.
mistake :
initialize, recall that broker[i] will be changed in every loop of choosing new source vertex
for ( int i = 1 ; i <= b_num; i ++ ) { // at beginning Q set = 1,2, ... b_num Q.insert(i); broker[i] = broker1[i];}