Description

In a highly modernized fishing village, inhabitants there make a living on fishery. Their major tools, fishing nets, are produced and fixed by computer. After catching fishes each time, together with plenty of fishes, they will bring back the shabby fishing nets, which might be full of leaks. Then they have to inspect those nets. If there exist large leaks, they have to repair them before launching out again.

Obviously, the smaller the leaks in the fishing nets are, the more fishes they will catch. So after coming back, those fishermen will input the information of the fishing nets into the computer to check whether the nets have leaks.

The checking principle is very simple: The computer regards each fishing net as a simple graph constructed by nodes and edges. In the graph, if any circle whose length (the number of edges) is larger than 3 must has at least one chord, the computer will output "Perfect" indicating that the fishnet has no leaks. Otherwise, "Imperfect" will be displayed and the computer will try to repair the net.

Note: A circle is a closed loop, which starts from one node, passes through other distinct nodes and back to the starting node. A chord is an edge, which connects two different nodes on the circle, but it does not belong to the set of edges on the circle.

 

Solution

PE*1 QvQ

Follow the output for each net with a blank line.

最大势算法 Maximum Cardinality Search

由n到1的顺序给点标号，设label[i]表示i与多少个已标号的点相连，每次选择label[i]最大的未标号点标号

判断序列是否为完美消除

设{v_i+1,...,v_n}中所有与v_i相邻的点依次为v_j1,...,v_jk。只需判断v_j1是否与v_j2,...,v_jk相邻即可。

#include
     
       #include
      
       #include
       
        #include
        
         #define MAXN 1005using namespace std;int n,m,Map[MAXN][MAXN],label[MAXN],num[MAXN],visited[MAXN];void work(){    for(int i=n;i>0;i--)    {        int u=0;        for(int j=1;j<=n;j++)        if(!visited[j]&&(!u||label[j]>label[u]))u=j;        visited[u]=1;num[i]=u;        for(int j=1;j<=n;j++)        {            if(Map[u][j]&&!visited[j])            label[j]++;        }    }}bool judge(){    for(int i=1;i