2-sat入门hdu1814

推荐 文件：http://wenku.baidu.com/view/afd6c436a32d7375a41780f2.html

http://wenku.baidu.com/view/0f96c3daa58da0116c1749bc.html

sat就是Satisfiability的缩写，就是适应满足的意思。2-sat的意思就是给你一些集合，每个集合最多有2个元素，然后给你一些条件，让你判断能否选出复合条件的。

文件中说的就是这个题。

意思什么的第一个文件说的已经很清楚了，这里就不在赘述了。

Peaceful Commission

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2795    Accepted Submission(s): 874

Problem Description

The Public Peace Commission should be legislated in Parliament of The Democratic Republic of Byteland according to The Very Important Law. Unfortunately one of the obstacles is the fact that some deputies do not get on with some others. 

The Commission has to fulfill the following conditions: 
1.Each party has exactly one representative in the Commission, 
2.If two deputies do not like each other, they cannot both belong to the Commission. 

Each party has exactly two deputies in the Parliament. All of them are numbered from 1 to 2n. Deputies with numbers 2i-1 and 2i belong to the i-th party . 

Task 
Write a program, which: 
1.reads from the text file SPO.IN the number of parties and the pairs of deputies that are not on friendly terms, 
2.decides whether it is possible to establish the Commission, and if so, proposes the list of members, 
3.writes the result in the text file SPO.OUT. 

 

Input

In the first line of the text file SPO.IN there are two non-negative integers n and m. They denote respectively: the number of parties, 1 <= n <= 8000, and the number of pairs of deputies, who do not like each other, 0 <= m <=2 0000. In each of the following m lines there is written one pair of integers a and b, 1 <= a < b <= 2n, separated by a single space. It means that the deputies a and b do not like each other. 
There are multiple test cases. Process to end of file. 

 

Output

The text file SPO.OUT should contain one word NIE (means NO in Polish), if the setting up of the Commission is impossible. In case when setting up of the Commission is possible the file SPO.OUT should contain n integers from the interval from 1 to 2n, written in the ascending order, indicating numbers of deputies who can form the Commission. Each of these numbers should be written in a separate line. If the Commission can be formed in various ways, your program may write mininum number sequence. 

 

Sample Input

   
   
   
   

    
    
    
    3 2
1 3
2 4
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    1
4
5
   
   
   
   
   
   
   
   


   
   
   
   

#include <cstdio>
#include <cstring>
#include <cctype>
#include <cmath>
#include <set>
#include <map>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <string>
#include <bitset>
#include <vector>
#include <iostream>
#include <algorithm>
#define max(a,b) ((a)>(b)?(a):(b))
#define mem(a,b) memset(a,b,sizeof(a))
#define For(a,l,r) for(int a=l;a<r;a++)
using namespace std;

typedef  long long LL;
const LL  mod = 9973;
const LL  inf=0x3f3f3f3f;
const double pi=acos(-1);
const int N=80020;//最大点数
const int M=200020;// 最大边数
struct Edge
{
	int to,next;
}ed[M];
int head[N],tot;
void init()
{
	tot=0;
	memset(head,-1,sizeof(head));
}
void add(int u,int v)
{
	ed[tot].to=v;
	ed[tot].next=head[u];
	head[u]=tot++;
}
bool vis[N];
int s[N],top;
bool dfs(int u)
{
	if(vis[u^1]) return false;
	if(vis[u]) return true;
	vis[u]=true;
	s[top++]=u;
	for(int i=head[u];i!=-1;i=ed[i].next)
		if(!dfs(ed[i].to))
			return false;
		return true;
}
bool tset(int n)
{
	memset(vis,0,sizeof(vis));
	for(int i=0;i<n;i+=2)
	{
		if(vis[i]||vis[i^1])
			continue;
		top=0;
		if(!dfs(i))
		{
			while(top) vis[s[--top]]=false;
			if(!dfs(i^1))return false;
		}
	}
	return true;
}
int main()
{
#ifdef LOCALHEI
    freopen("date.in","r",stdin);
    freopen("date.out","w",stdout);
#endif
    int n,m;
    int u,v;
    while(scanf("%d%d",&n,&m)==2)
    {
    	init();
    	while(m--)
    	{
    		scanf("%d%d",&u,&v);
    		u--;
    		v--;
    		add(u,v^1);
    		add(v,u^1);
    	}
    	if(tset(2*n))
    	{
    		for(int i=0;i<2*n;i++)
    			if(vis[i])
    				printf("%d\n",i+1);
    	}
    	else puts("NIE");

    }
   
    return 0;
}