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poj1125&zoj1082StockbrokerGrapevine(Floyd算法)

StockbrokerGrapevineTimeLimit:1000MSMemoryLimit:10000KDescriptionStockbrokersareknowntoo

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 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

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

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id="codeSnippet">

1: #include

2: #include

3: #include

4: #include

5: using namespace std;

6: const int INF=2147483647;

7: const int maxn=100;

8: 

9: int g[maxn+2][maxn+2];//储存人际关系

10: int n;

11: 

12: void floyd()

13: {

14: int i,j,k;

15: for(k=0; k

16: for(i=0; i

17: {

18: if(g[i][k]!=0)//i,k之间有路径

19: for(j=0; j

20: {

21: if(g[k][j]&&i!=j)//k,j之间有路径

22: if(g[i][j]==0||(g[i][j]>g[i][k]+g[k][j]))

23: g[i][j]=g[i][k]+g[k][j];

24: }

25: }

26: }

27: 

28: void show()//对本题无用,只是输出中间过程便于观察

29: {

30: int i,j;

31: for( i=0;i

32: {

33: for(j=0; j

34: printf("%d ",g[i][j]);

35: printf("\n");

36: }

37: }

38: 

39: void solve()

40: {

41: int i,j,a,time,m;

42: memset(g,0,sizeof(g));

43: for(i=0; i

44: scanf("%d",&m);

45: for(j=0;j

46: scanf("%d%d",&a,&time);

47: g[i][--a]=time;

48: }

49: }

50: floyd();

51: //show();

52: int mi=INF;

53: for(i=0;i

54: {

55: time=0;

56: for(j=0; j

57: if(i!=j)

58: {

59: if(g[i][j]==0){ time=INF; break; }

60: time=max(g[i][j],time);

61: }

62: if(mi>time){

63: a=i,mi=time;

64: }

65: }

66: if(mi"%d %d\n", ++a, mi);

67: else printf("disjoint\n");

68: }

69: 

70: int main()

71: {

72: //freopen("in.txt","r",stdin);

73: while(scanf("%d",&n)&&n)

74: solve();

75: return 0;

76: }


poj1125&zoj1082Stockbroker Grapevine(Floyd算法),布布扣,bubuko.com


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