HDU1814(2-SAT)

Peaceful Commission

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4180    Accepted Submission(s): 1395

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

 

染色法的2-SAT

//2017-08-28
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>

using namespace std;

const int N = 20010;
const int M = 100000;
const int INF = 0x3f3f3f3f;
int head[N], tot;
struct Edge{
    int to, next;
}edge[M];

void init(){
    tot = 0;
    memset(head, -1, sizeof(head));
}

void add_edge(int u, int v){
    edge[tot].to = v;
    edge[tot].next = head[u];
    head[u] = tot++;
}

bool book[N];
vector<int> vec;//保存dfs过程中经过的点

//input: u 顶点
//output: true 从u开始染色，不会出现NOT u和u染为同一种颜色; false dfs染色失败
bool dfs(int u){
    if(book[u^1])return false;//表示染到非u，染色失败
    if(book[u])return true;
    book[u] = true;
    vec.push_back(u);
    for(int i = head[u]; i != -1; i = edge[i].next){
        int v = edge[i].to;
        if(!dfs(v))
          return false;
    }
    return true;
}

//input：n 图的顶点数
//output：true 存在可行解; false 不存在可行解
bool twoSAT(int n){
    memset(book, 0, sizeof(book));
    for(int u = 0; u < n; u += 2){
        if(book[u] || book[u^1])continue;
        vec.clear();
        if(!dfs(u)){//如果选u不成功，把dfs过程中的点都从答案中删去
            for(int i = 0; i < vec.size(); i++)
                book[vec[i]] = 0;
            vec.clear();
            if(!dfs(u^1))return false;//如果选NOT u也不成功，说明不存在可行解
        }
    }
    return true;
}

int n, m;

int main()
{
    std::ios::sync_with_stdio(false);
    //freopen("inputG.txt", "r", stdin);
    while(cin>>n>>m){
        init();
        int u, v;
        while(m--){
            cin>>u>>v;
            u--; v--;
            add_edge(u, v^1);// u -> NOT v
            add_edge(v, u^1);// v -> NOT u
        }
        if(twoSAT(n<<1)){
            for(int i = 0; i < (n<<1); i++)//字典序输出解
                  if(book[i])
                      cout<<i+1<<endl;
        }else cout<<"NIE"<<endl;
    }
    return 0;
}