XMU 1050 Diffuse Secret 【最短路】

1050: Diffuse Secret

Time Limit: 500 MS  Memory Limit: 64 MB
Submit: 10  Solved: 8
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Description

　　A secret is good in one, better in two, ill in three and worse in four. 
　　----Hi, I know a secret, come on I will tell you. But you must promise that you won’t tell it to other guys. 
　　----OK, of course I won't. 
　　Do you think they will keep the secret? No no, just like most people, they share their secrets with their own friends. If everybody knows a secret, it will never be secret any more. 
　　For some reasons that know to all, TheBeet keep the salaries of employees as secret. But the employees don’t. At the beginning, everyone only knows how much he/she get. Everyday they share every secret they know with their friends (They don’t share the secret that they just heard today). For Example, A and B are good friends, B and C are good friends, A knows the secret a and secret d. B knows the secret b and secret c. C knows the secret e. After they share their secrets, A will know the secret abcd, B will know the secret abcde, C will know the secret bce. Today TheBeet increases every employee’s salary and he wonders when will be known to all employees.

Input

　　The input file contains several test cases. 
　　Each test case begins with two integers N, M 0 < N <= 100, 0 <= M <= (N * (N – 1)), N is the number of employee and M indicates there are M pairs of friend. 
　　The following M lines contain 2 integers each line describing the pair of friends. 
N = 0 and M = 0 indicates the end of input and should not be processed.

Output

　　For each test case, for each test case, output “Case #:” on the first line, '#' is the number of the test case. Then output a single integer as after such days, these secrets will never be secret. If there is one secret that some one will never know, you just output “Secret.” (Without quotes)

Sample Input

5 5
1 2
2 3
3 4
4 5
5 1
3 1
1 2
0 0

Sample Output

Case 1:
2
Case 2:
Secret.

HINT

Source

廈門大學第四屆程序設計競賽 第一次網絡預賽 by TheBeet @ btALT

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題目鏈接：

　　http://acm.xmu.edu.cn/JudgeOnline/problem.php?id=1050

題目大意：

　　每個人都有個祕密，他們每一天都會把自己的所有知道的祕密與他朋友分享

　　當天得到的祕密不會分享。問所有人都知道所有祕密時是第幾天。

題目思路：

　　【最短路】

　　考慮第i個人的祕密最遲到j，則祕密從i到j的傳播爲從i到j的最短路。

　　那麼所有人的祕密都到達最遠的人的時間就爲答案，即求所有人到所有人的最短路里的最大值。

　　用floyd即可。

/****************************************************
	
	Author : Coolxxx
	Copyright 2017 by Coolxxx. All rights reserved.
	BLOG : http://blog.csdn.net/u010568270
	
****************************************************/
#include<bits/stdc++.h>
#pragma comment(linker,"/STACK:1024000000,1024000000")
#define abs(a) ((a)>0?(a):(-(a)))
#define lowbit(a) (a&(-a))
#define sqr(a) ((a)*(a))
#define mem(a,b) memset(a,b,sizeof(a))
const double EPS=1e-8;
const int J=10000;
const int MOD=100000007;
const int MAX=0x7f7f7f7f;
const double PI=3.14159265358979323;
const int N=104;
using namespace std;
typedef long long LL;
double anss;
LL aans;
int cas,cass;
int n,m,lll,ans;
int e[N][N],f[N][N];
void floyd()
{
	int i,j,k;
	for(k=1;k<=n;k++)
	{
		for(i=1;i<=n;i++)
		{
			if(i==k)continue;
			for(j=1;j<=n;j++)
			{
				if(i==j || k==j)continue;
				f[i][j]=min(f[i][j],f[i][k]+f[k][j]);
			}
		}
	}
}
int main()
{
	#ifndef ONLINE_JUDGE
	freopen("1.txt","r",stdin);
//	freopen("2.txt","w",stdout);
	#endif
	int i,j,k,l;
	int x,y,z;
//	for(scanf("%d",&cass);cass;cass--)
//	for(scanf("%d",&cas),cass=1;cass<=cas;cass++)
//	while(~scanf("%s",s))
	while(~scanf("%d",&n))
	{
		scanf("%d",&m);
		if(!n && !m)break;
		ans=0;mem(f,0x01);
		printf("Case %d:\n",++cass);
		for(i=1;i<=m;i++)
		{
			scanf("%d%d",&x,&y);
			f[x][y]=f[y][x]=e[x][y]=e[y][x]=1;
		}
		floyd();
		for(i=1;i<=n;i++)
		{
			for(j=1;j<=n;j++)
			{
				if(i==j)continue;
				ans=max(ans,f[i][j]);
			}
		}
		if(ans==0x01010101)puts("Secret.");
		else printf("%d\n",ans);
	}
	return 0;
}
/*
//

//
*/