RT，Debug 调到自闭了……

思路就是一个人与其厌恶的人的同伴连边，跑强联通分量，如果在一个分量里就矛盾，否则输出方案。

代码里有适量注释，有人能帮帮忙么/kel

Code:

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
inline int read(){
	int x = 0, f = 1; char ch = getchar();
	while(ch < '0' || ch > '9'){ if(ch == '-') f = -1; ch = getchar(); }
	while(ch >= '0' && ch <= '9'){ x = x * 10 + (ch ^ 48); ch = getchar(); }
	return x * f;
}
const int N = 40010;
struct edge{
	int u, v, nxt;
}e[N << 1];
int head[N << 1], cnt, n, m, x, y, vis[N << 1], f[N << 1];
inline void add(int u, int v){
	e[++cnt].u = u;
	e[cnt].v = v;
	e[cnt].nxt = head[u];
	head[u] = cnt;
}
inline int anc(int x){
	return x == f[x] ? x : f[x] = anc(f[x]);
}
inline void dfs(int u){
	for(int i = head[u]; i; i = e[i].nxt){
		int v = e[i].v;
		if(!vis[v]){
			vis[v] = vis[u] + 1;
			dfs(v);
		}
		int fu = anc(u), fv = anc(v);
		if(vis[fv] > 0){
			if(vis[fu] < vis[fv]) f[fv] = fu;
			else f[fu] = fv;
		}
	}
	vis[u] = -1;
}//跑强联通分量 
int main(){
	n = read(); m = read();
	for(int i = 1; i <= m; ++i){
		x = read(); y = read();
		add(x, (y & 1) ? (y + 1) : (y - 1));
		add(y, (x & 1) ? (x + 1) : (x - 1));
	}//建图 
	for(int i = 1; i <= (n << 1); ++i) f[i] = i;//并查集初始化 
	for(int i = 1; i <= (n << 1); ++i){
		if(!vis[i]){
			vis[i] = 1;
			dfs(i);
		}
	}
	for(int i = 1; i <= n; ++i){
		if(anc(2 * i - 1) == anc(2 * i)) return puts("NIE") & 0;
	}//在一个强联通分量里，矛盾 
	for(int i = 1; i <= n; ++i){
		if(anc(2 * i - 1) < anc(2 * i)) printf("%d\n", 2 * i - 1);
		if(anc(2 * i - 1) > anc(2 * i)) printf("%d\n", 2 * i);
	}//输出方案 
	return 0;
}