hdu 1814 Peaceful Commission （2-SAT）

Peaceful Commission

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3355    Accepted Submission(s): 1101

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

 

Recommend

威士忌   |   We have carefully selected several similar problems for you:   1815  1816  3062  1811  1817 

 

Statistic |  Submit |  Discuss | Note

题解：2-SAT

如果x,y之间有边，那么x,y不能同时入选，也就是选了x必然要选y',选了y必然要选x'

初始时所有的标记都是-1.从编号小的节点开始尝试，将当前尝试节点x都标记为1，x'标记为0，然后对于他的后继节点进行标记，如果后继节点没有标记，则标记为1。如果标记是1，说明这个强连通分量遍历结束，并且可行。如果标记为0，则说明存在节点x,x'在同一个强连通分量中，发生冲突，当前尝试不合法，需要将标记撤回，从最开始节点的对立点继续进行尝试。

#include
#include
#include
#include
#include
#define N 50003
using namespace std;
int n,m,cnt;
int point[N],v[N],nxt[N],col[N],q[N],tot;
void add(int x,int y)
{
	tot++; nxt[tot]=point[x]; point[x]=tot; v[tot]=y;
}
bool dfs(int x)
{
	if (col[x]==1) return true;
	if (col[x]==0) return false;
	q[++cnt]=x;
	col[x]=1; col[x^1]=0;
	for (int i=point[x];i;i=nxt[i])
	 if (!dfs(v[i])) return false;
	return true;
}
bool solve()
{
	for (int i=0;i<2*n;i++) {
		cnt=0;
		if (!dfs(i)) {
			for (int j=1;j<=cnt;j++) {
				int t=q[j];
				col[t]=-1; col[t^1]=-1;
			}
			if (!dfs(i^1)) return false;
		}
	}
	return true;
}
int main()
{
	freopen("a.in","r",stdin);
	while (scanf("%d%d",&n,&m)!=EOF){
		tot=0;
		memset(col,-1,sizeof(col));
		memset(point,0,sizeof(point));
		for (int i=1;i<=m;i++) {
			int x,y; scanf("%d%d",&x,&y);
			x--; y--;
			add(x,y^1); add(y,x^1);
		}
		if (solve()){
			for (int i=0;i<2*n;i++)
			 if (col[i]==1) printf("%d\n",i+1);
			continue;
		}
		printf("NIE\n");
    }
}