HDU 3926 并查集 图同构简单判断 STL

给出两个图，问你是不是同构的...

直接通过并查集建图，暴力用SET判断下子节点个数就行了。

/** @Date    : 2017-09-22 16:13:42
  * @FileName: HDU 3926 并查集 图同构 连通分量.cpp
  * @Platform: Windows
  * @Author  : Lweleth ([e-mail])
  * @Link    : https://github.com/
  * @Version : $Id$
  */
#include <bits/stdc++.h>
#define LL long long
#define PII pair<int ,int>
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 1e4+20;
const double eps = 1e-8;

int fa[N];
int cnt[2][N],clr[2][N];

int find(int x)
{
	if(x != fa[x])
		fa[x] = find(fa[x]);
	return fa[x];
}

int join(int a, int b, int f)
{
	int x = find(a);
	int y = find(b);
	if(x != y)
	{
		fa[y] = x;
		cnt[f][x] += cnt[f][y];
		cnt[f][y] = 0;
		return 1;
	}
	clr[f][x] = 1;//环标记
	return 0;
}
int main()
{
	int T;
	cin >> T;
	int icase = 0;
	while(T--)
	{
		int n, m;
		set<PII >s[2];
		for(int i = 0; i < 2; i++)
		{
			cin >> n >> m;
			for(int j = 0; j <= n; j++)
				fa[j] = j, cnt[i][j] = 1, clr[i][j] = 0;
			for(int j = 0; j < m; j++)
			{
				int x, y;
				scanf("%d%d", &x, &y);
				join(x, y, i);
			}
			for(int j = 1; j <= n; j++)//环和非环分组，对照其父亲对于子节点个数
				s[i].insert(MP(clr[i][j], cnt[i][j]));
		}
		/*for(auto i = s[0].begin(), j = s[1].begin(); i != s[0].end(); i++, j++)
			cout << i->fi << j->fi << endl;*/

		printf("Case #%d: %s\n", ++icase, (s[0]==s[1])?"YES":"NO");
	}
    return 0;
}//STL大法好！！