Light OJ 1201 - A Perfect Murder

1201 - A Perfect Murder

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Time Limit: 2 second(s)Memory Limit: 32 MB

"Yes, I am the murderer. No doubt" I had to confess it in front of all. But wait, why I am confessing? Nobody wants to go to jail, neither do I. As you have suspected there is something fishy. So, let me explain a bit.

The murder was happened in 19th June, at 11:30 pm this year (2009) according to the medical report. So, I was asking the judge "Can you find the time 19th June 11:30 pm in Bangladesh?" The judge informed other reporters to find the time. But alas! There was no time - "2009, 19th June, 11:30 pm". So, the judge got a bit confused about my confession. So, I began to tell them, "The time the murder was happened, is not a valid time according to you. So, how can you claim that I am the murderer?"

And what happened next, you all know. I am in the streets again with a clean sheet.

But now I have planned to kill again. I have a list of N mosquitoes which are to be killed. But there is a small problem. If I kill a mosquito, all of his friends will be informed, so they will be prepared for my attack, thus they will be impossible to kill. But there is a surprising fact. That is if I denote them as a node and their friendship relations as edges, the graph becomes acyclic.

Now I am planning when and how to kill them (how to get rid of the law!) and you have to write a program that will help me to find the maximum number of mosquito I can kill. Don't worry too much, if anything goes wrong I will not mention your name, trust me!

Input

Input starts with an integer T (≤ 50), denoting the number of test cases.

Each case starts with a blank line and two integers N (1 ≤ N ≤ 1000) denoting the number of mosquito I want to kill and M denoting the number of friendship configurations. Each of the next Mlines contains two integers a and b denoting that a^th and b^th mosquitoes are friends. You can assume that (1 ≤ a, b ≤ N, a ≠ b) and each friendship relation is given only once. As I have already mentioned, you will not find any cycle in the relations.

Output

For each case, print the case number and the maximum number of mosquitoes I can kill considering the conditions described above.

Sample Input

Output for Sample Input

3

 

4 3

1 2

1 3

1 4

 

3 2

1 2

2 3

 

5 4

1 2

1 3

2 4

2 5

Case 1: 3

Case 2: 2

Case 3: 3

 

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PROBLEM SETTER: JANE ALAM JAN

代码：

#include<cstdio>#include<cstring>bool map[1020][1020];bool fafe[1010]; int shu[1010];int main(){int t;scanf("%d",&t);for (int ca=1;ca<=t;ca++){int n,m;scanf("%d%d",&n,&m);memset(map,false,sizeof(map));memset(fafe,true,sizeof(fafe));memset(shu,0,sizeof(shu));int a,b;for (int i=0;i<m;i++){scanf("%d%d",&a,&b);shu[a]++;shu[b]++;map[a][b]=map[b][a]=true;}int s=0;while (1){bool lp=false;int ii,mi=100000;for (int i=1;i<=n;i++)if (fafe[i]&&shu[i]<mi){lp=true;ii=i;mi=shu[i];}if (lp==false)break;s++;fafe[ii]=false;for (int i=1;i<=n;i++)if (fafe[i]&&map[i][ii]){fafe[i]=false;for (int j=1;j<=n;j++)if (fafe[j]&&map[i][j])shu[j]--;}}printf("Case %d: %d\n",ca,s);}return 0;}