USTC 1119 graph 图的同构

USTC 1119

图的同构的严格定义可以参考离散数学:The simple graphs G1=(V1,E1) and G2=(V2,E2)are isomorphic if there exists a one to one and onto function f from V1 to V2 with the property that a and b are adjacent in G1 if and only if f(a)and f(b) are adjacent G2 , for all a and b in V1.

用哈希映射的方法可以解决,这个题的数据规模比较小，应该可以解决规模更大的题目。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn=30,maxm=300,K=10,A=7,B=3,C=4,P=10007,D=35;
int a[maxm][2],co[maxn],f[maxn],tf[maxn],n,m,cot[maxn];
void graph_hash()
{
 int q,w,e;
 for(int i=1;i<=n;i++)
 	{
 	 for( q=1;q<=n;q++)f[q]=1;
 	 for(int z=0;z<K;z++)
 	 	{
 	 	 memcpy(tf,f,sizeof(f));
		 for(q=1;q<=n;q++)f[q]*=A;
		 for(q=0;q<m;q++)
		 	{
		 	 f[a[q][0]]+=tf[a[q][1]]*B;
		 	 f[a[q][1]]+=tf[a[q][0]]*C;
			}
		 f[i]+=D;
		 for(q=1;q<=n;q++)f[q]%=P;
		}
	 co[i]=f[i];
	}
sort(co+1,co+1+n);
}
int main()
{
 int kt;scanf("%d",&kt);
 for(int ii=0;ii<kt;ii++)
 	{
 	 scanf("%d%d",&n,&m);
 	 for(int i=0;i<m;i++)
 	 	{
		  scanf("%d%d",&a[i][0],&a[i][1]);
		  a[i+m][0]=a[i][1];
		  a[i+m][1]=a[i][0];
		}
	 m=m*2;
	 graph_hash();
	 memcpy(cot,co,sizeof(co));
	 m=m/2;
	 for(int i=0;i<m;i++)
 	 	{
		  scanf("%d%d",&a[i][0],&a[i][1]);
		  a[i+m][0]=a[i][1];
		  a[i+m][1]=a[i][0];
		}
	 m=m*2;
	graph_hash();
	bool ans=true;
	for(int i=1;i<=n;i++)
		{
		 if(cot[i]!=co[i]){ans=false;break;}
		}
	 if(ans==false){printf("different\n");}
	 	else printf("same\n");
	}
 return 0;
}