hdu 1814 Peaceful Commission （2-sat 输出字典序最小的路径）

Peaceful Commission

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

Total Submission(s): 1948    Accepted Submission(s): 560

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：

如今有n个党派,每一个党派有2个代表,我们须要从每一个党派中选一个代表出来,构成一个n个人的立法委员会.可是可能有一些代表互相讨厌,所以他们不能同一时候出如今立法委员会中.如今问你是否存在一个合理的方案,且输出全部可能立法委员会的最小字典序结果.

思路：

裸的2-sat题。主要是练习一下输出路径，依据给的矛盾关系建图，最后依据mark是否标记来输出路径。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#define maxn 8005
#define MAXN 400005
#define OO (1<<31)-1
#define mod 1000000007
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
typedef long long ll;
using namespace std;

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<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 xval,int y,int yval)
    {
        x=x*2+xval;
        y=y*2+yval;
        g[x^1].push_back(y);
        g[y^1].push_back(x);
    }
    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 ;
    }
};

int n,m,sum;
int age[maxn];
TwoSAT ts;

int main()
{
    int i,j;
    while(~scanf("%d%d",&n,&m))
    {
        ts.init(n);
        int u,v,x,y;
        for(i=1; i<=m; i++)
        {
            scanf("%d%d",&u,&v);
            u--;
            v--;
            ts.g[u].push_back(v^1);
            ts.g[v].push_back(u^1);
        }
        if(ts.solve())
        {
            for(i=0; i<n*2; i+=2)
            {
                if(ts.mark[i]) printf("%d\n",i+1);
                else printf("%d\n",i+2);
            }
        }
        else puts("NIE");
    }
    return 0;
}

版权声明：本文博客原创文章，博客，未经同意，不得转载。