Peaceful Commission

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 79 Accepted Submission(s): 38

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

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威士忌

#include<bits/stdc++.h>
using namespace std;
/*********************************************2-SAT模板*********************************************/
const int maxn=8000+10;
struct TwoSAT
{
    int n;//原始图的节点数(未翻倍)
    vector<int> G[maxn*2];//G[i]==j表示如果mark[i]=true,那么mark[j]也要=true
    bool mark[maxn*2];//标记
    int S[maxn*2],c;//S和c用来记录一次dfs遍历的所有节点编号

    void init(int n)
    {
        this->n=n;
        for(int i=0;i<2*n;i++) G[i].clear();
        memset(mark,0,sizeof(mark));
    }

    //加入(x,xval)或(y,yval)条件
    //xval=0表示假，yval=1表示真
    void add_clause(int x,int y)
    {
        G[y].push_back(x^1);
        G[x].push_back(y^1);
    }

    //从x执行dfs遍历,途径的所有点都标记
    //如果不能标记,那么返回false
    bool dfs(int x)
    {
        if(mark[x^1]) return false;//这两句的位置不能调换
        if(mark[x]) return true;
        mark[x]=true;
        S[c++]=x;
        for(int i=0;i<G[x].size();i++)
            if(!dfs(G[x][i])) return false;
        return true;
    }

    //判断当前2-SAT问题是否有解
    bool solve()
    {
        for(int i=0;i<2*n;i+=2)
        if(!mark[i] && !mark[i+1])//如果这个点没有进行标记
        {
            c=0;
            if(!dfs(i))
            {
                while(c>0) mark[S[--c]]=false;
                if(!dfs(i+1)) return false;
            }
        }
        return true;
    }
    void print()  
    {  
        if(!solve()) printf("NIE
");  
        else  
        {  
            for(int i=0;i<2*n;i++)if(mark[i])  
                printf("%d
",i+1);  
        }  
    }  
}sat;
/*********************************************2-SAT模板*********************************************/
int n,m;
int a,b;
int main(){
    // freopen("in.txt","r",stdin);
    while(scanf("%d%d",&n,&m)==2){
        sat.init(n);
        for(int i=0;i<m;i++){
            scanf("%d%d",&a,&b);
            a--;b--;
            sat.add_clause(a,b);
        }
        sat.print();
    }
    return 0;
}