Hdu1869 Floyd)

Question link: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 1869

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Problem description in 1967, the famous American sociologist Stanley milgranm proposed a famous hypothesis named "small world phenomenon, no matter what two strangers, at most six people are separated, that is, only six people can be associated with each other, therefore, his theory is also called six degrees of separation ). Despite the frequent fulfillment of milgalham's theory, many scientists have always been interested in it. However, over the past 30 years, it has never been rigorously proved, it is just a legend.

Lele is very interested in this theory, so he investigates N people in HDU. He has known the relationship between them. Now, please help him verify if "Six-Degree separation" is true. Input this question includes multiple groups of samples. Please process the results until the end of the file.
For each benchmark test, the first line contains two integers n, m (0 <n <200, 0 <m <), representing the number of people in the HDU respectively (these people are compiled into 0 ~ N-1 number) and the relationship between them.
Next, there are m rows. Each row has two integers, A and B (0 <= A, B <n), which indicate that the people numbered A and B in HDU know each other.
In addition to the M group relationship, other casual people do not know each other.
Output for each group of samples, if the data conforms to the "Six Degree separation" theory, "yes" is output in a row; otherwise, "no" is output ". Sample Input

 
8 70 11 22 33 44 55 66 78 80 11 22 33 44 55 66 77 0
 

Sample output

 
Yesyes
 

Authorlinle source2008 select-a warm-up competition

Idea: we only need to set the value of the two people we know to 1, and then Floyd again. Suppose there is a distance between two people greater than 7 (six people form seven segments ), then output "no" and vice versa "yes ".
The Code is as follows:

 
# Include <cstdio> # include <cstring> # define INF 99999999 # define M 217int n, m, a, B, X; int dis [m] [m]; int min (int A, int B) {int m; M = A <B? A: B; return m;} void Floyd () {for (int K = 0; k <n; k ++) {for (INT I = 0; I <N; I ++) {for (Int J = 0; j <n; j ++) {dis [I] [J] = min (DIS [I] [J], dis [I] [k] + dis [k] [J]) ;}}} void Init () {for (INT I = 0; I <N; I ++) {for (Int J = 0; j <n; j ++) {if (I = J) dis [I] [J] = 0; // This step must be classified as zero elsedis [I] [J] = inf ;}}int main () {int I, j; int cont [m]; while (~ Scanf ("% d", & N, & M) {Init (); memset (cont, 0, sizeof (cont); for (I = 0; I <m; I ++) {scanf ("% d", & A, & B ); dis [a] [B] = dis [B] [a] = 1;} Floyd (); int flag = 0; for (I = 0; I <N; I ++) {for (j = 0; j <n; j ++) {If (DIS [I] [J]> 7) {flag = 1; break ;}}if (FLAG) printf ("NO \ n"); elseprintf ("Yes \ n");} return 0 ;}
 

hdu1869 Floyd