HOJ 1917 POI 2001 Peaceful Commission 2-SAT问题
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
传送门
#include
#include
#include
#include
using namespace std;
int read(){
int x=0,f=1;char ch=getchar();
while (ch<'0' || ch>'9'){if (ch=='-') f=-1;ch=getchar();}
while (ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
const int
N=16010,
M=40005;
int n,m,scc,Ecnt,Time,top;
int DFN[N],LOW[N],stk[N],num[N];
int X[M<<1],Y[M<<1],TY[N];
int into[N],color[N];
bool vis[N],instack[N];
queueq;
struct Edge{
int next,to;
}E[M<<1];int head[N];
int ty(int x){
if (x&1) return x+1;
else return x-1;
}
void add(int u,int v){
E[++Ecnt].next=head[u];
E[Ecnt].to=v;
head[u]=Ecnt;
}
void tarjan(int u){
DFN[u]=LOW[u]=++Time;
stk[++top]=u;
instack[u]=1;
for (int i=head[u];i;i=E[i].next){
int j=E[i].to;
if (!DFN[j]){
tarjan(j);
LOW[u]=min(LOW[u],LOW[j]);
} else
if (instack[j]) LOW[u]=min(LOW[u],DFN[j]);
}
if (DFN[u]==LOW[u]){
int now=-1;scc++;
for (;now!=u;top--){
now=stk[top];
num[now]=scc;
instack[now]=0;
}
}
}
bool ok(){
for (int i=1;i<=n;i+=2)
if (num[i]==num[i+1]) return 0;
return 1;
}
void suodian(){
scc=0;
memset(DFN,0,sizeof(DFN));
memset(LOW,0,sizeof(LOW));
for (int i=1;i<=n;i++)
if (!DFN[i]) Time=top=0,tarjan(i);
Ecnt=0;
memset(head,0,sizeof(head));
memset(into,0,sizeof(into));
for (int i=1;i<=(m<<1);i++)
if (num[X[i]]!=num[Y[i]])
add(num[Y[i]],num[X[i]]),into[num[X[i]]]++;
for (int i=1;i<=n;i+=2)
TY[num[i]]=num[i+1],TY[num[i+1]]=num[i];
}
void toposort(){
memset(color,0,sizeof(color));
for (int i=1;i<=scc;i++)
if (!into[i]) q.push(i);
while (!q.empty()){
int u=q.front();
q.pop();
if (!color[u]) color[u]=1,color[TY[u]]=2;
for (int i=head[u];i;i=E[i].next){
int v=E[i].to;
into[v]--;
if (!into[v]) q.push(v);
}
}
}
void print(){
for (int i=1;i<=n;i++)
if (color[num[i]]==1) printf("%d\n",i);
}
void solve(){
int x,y;n<<=1;
memset(head,0,sizeof(head));
for (int i=1;i<=m;i++){
x=read(),y=read();
add(x,ty(y)),add(y,ty(x));
X[(i<<1)-1]=x,Y[(i<<1)-1]=ty(y);
X[i<<1]=y,Y[i<<1]=ty(x);
}
suodian();
if (!ok()){puts("NIE");return;}
toposort();
print();
}
int main(){
while (~scanf("%d%d",&n,&m)) solve();
return 0;
}