LOJ 1201 - A Perfect Murder（二分匹配 最大独立集）

题目链接：http://lightoj.com/volume_showproblem.php?problem=1201

"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 M lines 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

PS:

二分图最大独立集=顶点数-二分图最大匹配

代码如下：

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;//顶点编号从1开始的#define MAXN 1017int LN, RN;//L,R数目int g[MAXN][MAXN], linker[MAXN];bool used[MAXN];int dfs(int L)//从左边开始找增广路径{    int R;    for(R = 1; R <= RN; R++)    {        if(g[L][R]!=0 && !used[R])        {            //找增广路，反向            used[R]=true;            if(linker[R] == -1 || dfs(linker[R]))            {                linker[R]=L;                return 1;            }        }    }    return 0;//这个不要忘了，经常忘记这句}int hungary(){    int res = 0 ;    int L;    memset(linker,-1,sizeof(linker));    for( L = 1; L <= LN; L++)    {        memset(used,0,sizeof(used));        if(dfs(L) != 0)            res++;    }    return res;}int main(){    int t;    int n, m;    int cas = 0;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        memset(g,0,sizeof(g));        for(int i = 1; i <= m; i++)        {            int x, y;            scanf("%d%d",&x,&y);            g[x][y] = 1;            g[y][x] = 1;        }        LN = n;        RN = n;        int ans = hungary();//最大匹配数        printf("Case %d: %d\n",++cas,n-ans/2);    }    return 0 ;}