EOJ 1837小强的烦恼

思路

使用带权并查集处理这个题目，根据敌人的敌人是朋友的原则，xy的边权为偶数，则xy是朋友，否则是敌人。如果xy不在同一个并查集里面，则暂时不确定。

code

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 10;
int fa[MAXN];
int edge[MAXN];
inline int find(int x)
{
	if(x == fa[x]) return x;
	int tf = fa[x];
	fa[x] = find(fa[x]);
	edge[x] = (edge[x] + edge[tf]) % 2;
	return fa[x];
}
inline void merge(int x, int y)
{
	if(x == y) return;
	edge[x] = edge[y] + 1;	
	x = find(x);
	y = find(y);
	fa[x] = y;
}
int n, m;
inline void sov()
{
	cin >> n >> m;
	for(int i = 1; i <= n; ++i)
		fa[i] = i;
	for(int i = 1; i <= n; ++i) 
		edge[i] = 0;
	while(m--)
	{
		char c;
		int x, y;
		cin >> c >> x >> y;
		if(c == 'A')
		{
			merge(x, y);
		}
		else if(c == 'Q')
		{
			int nx = find(x);
			int ny = find(y);
			if(nx != ny)
			{
				cout << "Not sure yet.\n";
			}
			else
			{
				int d = edge[x] + edge[y];
//				cerr << nx << " " << edge[x] << " " << edge[y] << endl;
				if(d % 2 == 0)
				{
					cout << "In the same gang.\n";	
				}
				else
				{
					cout << "In different gangs.\n";
				}	
			}
		} 
	}
}
int main()
{
	int t;
	cin >> t;
	while(t--) sov();
}