http://acm.pku.edu.cn/JudgeOnline/problem?id=1125
Stockbroker Grapevine
Time Limit:1000MS Memory Limit:10000K
Total Submit:2602 Accepted:1503
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 spread the rumours in the fastest possible way.
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
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.
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 Output
3 23 10
Source
Southern African 2001
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Floyd-Warshall算法是解決任意兩點(diǎn)間的最短路徑的一種算法，可以正確處理有向圖（Directed Graph）或負(fù)數(shù)的代價(jià)(negtive cost)的最短路徑問(wèn)題。Floyd-Warshall算法的時(shí)間復(fù)雜度為<math>O(N^3)</math>。
Floyd-Warshall算法的描述如下：
for k ← 1 to n do
for i ← 1 to n do
for j ← 1 to n do
if (<math>D_{i,k} + D_{k,j} < D_{i,j}</math>) then
<math>D_{i,j}</math> ← <math>D_{i,k} + D_{k,j}</math>;
其中<math>D_{i,j}</math>表示由點(diǎn)<math>i</math>到點(diǎn)<math>j</math>的代價(jià)(cost)，當(dāng)<math>D_{i,j}</math>為 ∞ 表示兩點(diǎn)之間沒(méi)有任何連接（Disconnected）。
Floyd算法也可以說(shuō)是動(dòng)態(tài)規(guī)劃。
Source
Problem Id:1125 User Id:lnmm
Memory:84K Time:0MS
Language:C++ Result:Accepted
1
#include"stdio.h"
2int a[101][101];
3int i,j,k=0;
4int min;
5int max[101];
6int T;
7int n,m,temp,to;
8int flag;
9void main()
10
{
11
while(scanf("%d",&n)&&n!=0) //讀入一個(gè)set的人數(shù)
12
{
13 for(i=1;i<=n;i++)
14 for(j=1;j<=n;j++)
15 a[i][j]=32767;
16 for(i=1;i<=n;i++)
17 a[i][i]=0; //初識(shí)化該set的矩陣
18 for(i=1;i<=n;i++) //讀入一個(gè)set的數(shù)據(jù)
19 {
20 scanf("%d",&m);
21 for(j=1;j<=m;j++)
22 {
23 scanf("%d %d",&to,&temp);
24 a[i][to]=temp;
25
}
26 }
27 for(k=1;k<=n;k++) //弗洛伊德算法
28 for(i=1;i<=n;i++)
29 for(j=1;j<=n;j++)
30 {
31 if(a[i][k]!=32767 && a[k][j]!=32767 && a[i][j]>a[i][k]+a[k][j])
32 a[i][j]=a[i][k]+a[k][j];
33 }
34
35
36 flag=0;
37 for(i=1;i<=n;i++) //求出從i人開(kāi)始，謠言傳遞需要的時(shí)間
38 { max[i]=0;
39 for(j=1;j<=n;j++)
40 {
41 if(max[i]<a[i][j])max[i]=a[i][j];
42 }
43
44 }
45
46 min=32767; //計(jì)算最小謠言時(shí)間
47 for(i=1;i<=n;i++)
48 if(min>max[i])
49 {min=max[i];
50 k=i;
51 }
52 if(min==32767)printf("disjoint.\n");
53 else printf("%d %d\n",k,min);
54
55
56}
57
58
}