hdoj 1814 Peaceful Commission 【2-sat暴力染色 求解字典序最小解】

Peaceful Commission

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

Total Submission(s): 2343    Accepted Submission(s): 696

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

 

题意：

N个党派要成立一个委员会，此委员会必须满足下列条件：

• 每个党派都在委员会中恰有1个代表，
• 如果2个代表彼此厌恶，则他们不能都属于委员会。

每个党在议会中有2个代表。代表从1编号到2n。 编号为2i-1和2i的代表属于第I个党派。

现给出M个矛盾关系，问你该委员会能否创立？若不能输出NIE，若能够创立输出字典序最小的解。

dfs暴力：本想改下实现过程，结果一直WA。。。

#include 
#include 
#include 
#include 
#include 
#include 
#define MAXN 40000+10
#define MAXM 200000+10
#define INF 1000000
using namespace std;
struct Edge
{
	int from, to, next;
}edge[MAXM];
int head[MAXN], edgenum;
stack S;
bool mark[MAXN];//标记是否为真 
int N, M;
void init()
{
	edgenum = 0;
	memset(head, -1, sizeof(head));
	memset(mark, false, sizeof(mark));
} 
void addEdge(int u, int v)
{
	Edge E = {u, v, head[u]};
	edge[edgenum] = E;
	head[u] = edgenum++; 
}
void getMap()
{
	int a, b;
	while(M--)
	{
		scanf("%d%d", &a, &b);
		a--, b--;
		addEdge(a, b ^ 1);
		addEdge(b, a ^ 1);
	}
}
bool dfs(int u)
{
	if(mark[u ^ 1]) return false;
	if(mark[u]) return true;
	mark[u] = true;
	S.push(u);
	for(int i = head[u]; i != -1; i = edge[i].next)
	{
		int v = edge[i].to;
		if(!dfs(v)) return false;
	}
	return true;
}
bool solve()
{
	for(int i = 0; i < 2*N; i+=2)//共N组 
	{
		if(!mark[i] && !mark[i ^ 1])//还没有判定 
		{
			while(!S.empty())//用STL的话 注意清空栈 
			{
				S.pop();
			}
			if(!dfs(i))
			{
				while(!S.empty())
				{
					int v = S.top();
					S.pop();
					mark[v] = false;
				}
				if(!dfs(i ^ 1))//矛盾 必无解 
				return false;
			}
		}
	}
	return true;
}
int main()
{
	while(scanf("%d%d", &N, &M) != EOF)
	{
		init();
		getMap();
		if(solve())
		{
			for(int i = 0; i < 2*N; i++)
			if(mark[i]) printf("%d\n", i+1);
		}
		else
		printf("NIE\n");
	}
}