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:
- Each party has exactly one representative in the Commission,
- 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:
- reads from the text file SPO.IN the number of parties and the pairs of deputies that are not on friendly terms,
- decides whether it is possible to establish the Commission, and if so, proposes the list of members,
- 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 any of them.
Sample Input
3 2
1 3
2 4
Sample Output
1
4
5
最近看了2篇關于2-SAT問題的IOI論文，對2-SAT問題的O(m)時間復雜度的解法也有了一定的了解，找了道POI 2001的題來做，在WA了無數(shù)次之后終于過了，跑了1.44s，效率還可以。
2篇論文分別是<<由對稱性解2-SAT問題>>和<<2-SAT解法淺析>>。
//2-SAT問題
//求出所有強連通分量,如果有矛盾點同處于一個連通分量則無解
//縮點，將原圖反向建立DAG圖
//按拓撲排序著色，找一個未著色點x，染成紅色
//將與x矛盾的頂點及其子孫染為藍色
//直到所有頂點均被染色，紅色即為2-SAT的一組解
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
const int MAXN = 16010;//2*8000
char color[MAXN];//染色
bool visit[MAXN];
queue<int> q1,q2;
vector< vector<int> > adj; //原圖
vector< vector<int> > radj;//逆向圖
vector< vector<int> > dag; //縮點后的逆向DAG圖
int n,m,cnt,id[MAXN],order[MAXN],ind[MAXN];//強連通分量，訪問順序，入度
void dfs(int u)
{
visit[u]=true;
int i,len=adj[u].size();
for(i=0;i<len;i++)
if(!visit[adj[u][i]])
dfs(adj[u][i]);
order[cnt++]=u;
}
void rdfs(int u){
visit[u]=true;
id[u]=cnt;
int i,len=radj[u].size();
for(i=0;i<len;i++)
if(!visit[radj[u][i]])
rdfs(radj[u][i]);
}
void korasaju(){
int i;
memset(visit,false,sizeof(visit));
for(cnt=0,i=1;i<=2*n;i++)
if(!visit[i]) dfs(i);
memset(id,0,sizeof(id));
memset(visit,false,sizeof(visit));
for(cnt=0,i=2*n-1;i>=0;i--)
if(!visit[order[i]])
cnt++,rdfs(order[i]);
}
bool solvable(){
for(int i=1;i<=n;i++)
if(id[2*i-1]==id[2*i])
return false;
return true;
}
void topsort(){
int i,j,len,now,p,pid;
while(!q1.empty()){
now=q1.front();
q1.pop();
if(color[now]!=0) continue ;
color[now]='R';
ind[now]=-1;
for(i=1;i<=2*n;i++){
if(id[i]==now){
p=(i%2)?i+1:i-1;
pid=id[p];
q2.push(pid);
while(!q2.empty()){
pid=q2.front();
q2.pop();
if(color[pid]=='B') continue ;
color[pid]='B';
int len=dag[pid].size();
for(j=0;j<len;j++)
q2.push(dag[pid][j]);
}
}
}
len=dag[now].size();
for(i=0;i<len;i++){
ind[dag[now][i]]--;
if(ind[dag[now][i]]==0) q1.push(dag[now][i]);
}
}
}
int main(){
int i,j,x,y,xx,yy,len;
while(scanf("%d %d",&n,&m)!=EOF){
adj.assign(2*n+1,vector<int>());
radj.assign(2*n+1,vector<int>());
for(i=0;i<m;i++){
scanf("%d %d",&x,&y);
xx=(x%2)?x+1:x-1;
yy=(y%2)?y+1:y-1;
adj[x].push_back(yy);
adj[y].push_back(xx);
radj[yy].push_back(x);
radj[xx].push_back(y);
}
korasaju();
if(!solvable()) puts("NIE");
else{
dag.assign(cnt+1,vector<int>());
memset(ind,0,sizeof(ind));
memset(color,0,sizeof(color));
for(i=1;i<=2*n;i++){
len=adj[i].size();
for(j=0;j<len;j++)
if(id[i]!=id[adj[i][j]]){
dag[id[adj[i][j]]].push_back(id[i]);
ind[id[i]]++;
}
}
for(i=1;i<=cnt;i++)
if(ind[i]==0) q1.push(i);
topsort();
for(i=1;i<=n;i++){
if(color[id[2*i-1]]=='R') printf("%d\n",2*i-1);
else printf("%d\n",2*i);
}
}
}
return 0;
}