Poj1125 (stockbroker grapevine)

Question:

The question will be given to you by N brokers. The example shows the time required for communication and dissemination between the first brokers and other brokers, ask which broker to start spreading the rumors as quickly as possible and calculate the time required. The time here is the least time spent in a shortest path. If an economic person does not output "disjoint" in the shortest path, (Data Shui)

 

Solution:

The shortest path algorithm dijstra loops through every broker to find the answer that requires the least time path, that is, the question.

 

Code:

# Include <algorithm> # include <iostream> # include <sstream> # include <cstdlib> # include <cstring> # include <cstdio> # include <string> # include <bitset> # include <vector> # include <queue> # include <stack> # include <cmath> # include <list> // # include <map> # include <set> using namespace std; /*************************************** /# define ll long # define int64 _ int64 /*************************** * ***********/CONS T int inf = 0x7f7f7f7f; const double EPS = 1e-8; const double PIE = ACOs (-1.0); const int d1x [] = {0,-, 1 }; const int d1y [] = {-,}; const int d2x [] = {0,-, 1}; const int d2y [] =, -}; const int FX [] = {-1,-1,-,}; const int FY [] = {-, 1, -1, 0, 1 }; /*************************************** /void openfile () {freopen ("data. in "," rb ", stdin); freopen (" data. out "," WB ", stdout );}/**************** * ***** Gorgeous split line, the template section is *****************/INT map [500] [500]; int lowcost [500]; int vis [500]; int N; int CE; int maax; void dijstra (INT Sta) {int I, j, k = 0; int CNT = 0; int min; for (I = 1; I <= N; I ++) {lowcost [I] = map [sta] [I]; vis [I] = 0 ;} vis [sta] = 1; // lowcost [sta] = 0; for (I = 1; I <n; I ++) {min = inf; for (j = 1; j <= N; j ++) {If (! Vis [J] & lowcost [J] <min) {min = lowcost [J]; k = J ;}} if (k) CNT ++; if (CNT = N-1) if (Min <maax) {maax = min; Ce = sta;} vis [k] = 1; for (j = 1; j <= N; j ++) if (! Vis [J] & lowcost [J]> lowcost [k] + map [k] [J]) lowcost [J] = lowcost [k] + map [k] [J];} int main () {While (scanf ("% d", & N) & N) {int I, j; for (I = 0; I <500; I ++) for (j = 0; j <500; j ++) map [I] [J] = inf; int A; int U, cost; for (I = 1; I <= N; I ++) {scanf ("% d", & A); For (j = 0; j <A; j ++) {scanf ("% d", & U, & cost); map [I] [u] = Cost ;}} maax = inf; for (I = 1; I <= N; I ++) dijstra (I ); printf ("% d \ n", Ce, maax) ;}return 0 ;}// disjoint is not judged. You can add them by yourself.

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