hdu1814Peaceful Commission(2-SAT)

Peaceful Commission

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3556    Accepted Submission(s): 1173

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

 

 

Source

POI 2001

 

 

/*
最小字典序 
直接暴力枚举DFS，首先将所有的点都置为未染色，然后从第一个点开始DFS染色，我们先尝试将i染
成1(答案中的颜色)，将~i染成2，然后dfs i的所有后继并染色，如果对于后继j没有染色，那么将j然
后为1，~j染成2。如果后继j已经被染成2，则说明不能选则i，如果j已经染成1，则说明可以。
那么这些后继就可以被选择。如果选择i的时候失败了，那么必定要选择~i,如果也失败，则说明无解。否则
按次序选取下一个未被染色的点。时间复杂度O(nm)。
*/
#include<iostream> 
#include<cstdio>
#include<cstring>
#define maxn 16001

using namespace std;
int col[maxn],head[maxn],ans[maxn];
int a,b,n,m,num,tot,cnt;
struct node
{
    int u,v,next;
}e[maxn<<2];

inline void add(int u,int v)
{
    e[++num].u=u;
    e[num].v=v;
    e[num].next=head[u];
    head[u]=num;
}

bool dfs(int u)
{
    if(col[u]==1) return true;
    if(col[u]==2) return false;
    col[u]=1;col[u^1]=2;ans[cnt++]=u;
    for(int i=head[u];i;i=e[i].next)
    {
        int v=e[i].v;
        if(!dfs(v)) return false;
    }
    return true;
}

bool solve()
{
    memset(col,0,sizeof col);
    for(int i=0;i<n;i++)
    {
        if(col[i]) continue;
        cnt=0;
        if(!dfs(i))
        {
            for(int j=0;j<cnt;j++)
            {
                col[ans[j]]=0;
                col[ans[j]^1]=0;
            }
            if(!dfs(i^1))return false;
        }
    }
    return true;
}

int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        n<<=1;num=1;
        memset(head,0,sizeof head);
        while(m--)
        {
            scanf("%d%d",&a,&b);
            a--;b--;
            add(a,b^1);add(b,a^1);
        }
        if(solve())
        {
            for (int i=0;i<n;i++)
              if(col[i]==1)
                printf("%d\n",i+1);
        }
        else printf("NIE\n");
    }
    return 0;
}