HDU1814 2-sat 模板

Peaceful Commission

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3378    Accepted Submission(s): 1110

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

题意：

有2*n个人，第i个和第i+1个人属于一组，有m对人互相讨厌（不会是同一组的），问能否让每一组都有且只有一个人参加派对且派对上没有相互讨厌的人。字典序输出参加的人的编号。

代码：

//2-sat模板，白书323页。点的编号从0开始
//输出字典序最小的结果，模板跑出来就是字典序最小的。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
const int maxn=8003;
/********************** 2-sat模板 **********************/
struct Twosat{
    int n;
    vector<int> g[maxn*2];
    bool mark[maxn*2];
    int s[maxn*2],c;
    bool dfs(int x){
        if(mark[x^1]) return false;
        if(mark[x]) return true;
        mark[x]=true;
        s[c++]=x;
        for(int i=0;i<(int)g[x].size();i++)
            if(!dfs(g[x][i])) return false;
        return true;
    }
    void init(int n){
        this->n=n;
        for(int i=0;i<n*2;i++) g[i].clear();
        memset(mark,0,sizeof(mark));
    }
    void add_clause(int x,int y){//这个函数随题意变化
        g[x].push_back(y^1);//选了x就必须选y^1
        g[y].push_back(x^1);
    }
    bool solve(){
        for(int i=0;i<n*2;i+=2)
            if(!mark[i]&&!mark[i+1]){
                c=0;
                if(!dfs(i)){
                    while(c>0) mark[s[--c]]=false;
                    if(!dfs(i+1)) return false;
                }
            }
        return true;
    }
};
/*********************** 2-sat模板 ************************/
int main(){
    int n,m,a,b;
    Twosat solver;
    while(~scanf("%d%d",&n,&m)){
        solver.init(n);
        for(int h=0;h<m;h++){
            scanf("%d%d",&a,&b);//a,b不能同时选
            a--;b--;
            solver.add_clause(a,b);
        }
        if(solver.solve()){
            for(int i=0;i<n*2;i++)
                if(solver.mark[i]) printf("%d
",i+1);
        }
        else printf("NIE
");
    }
    return 0;
}