【2-sat】uva1391

输出方案采用了刘大师的写法，直接给每个强连通分量标号，每个人只要选属于较小编号的点即可。

#include 
#include 
#include 
#include 
#include 
using namespace std;
int n,m,ss,w_time,top,bj;
int tail[500000],next[2000000],sora[2000000];
int rel[500000],low[500000],st[500000],v[500000],b[500000],col[500000];
int age[500000],rank[500000];
void origin()
{
	ss=n+n;
	for (int i=1;i<=n+n;i++) tail[i]=i,next[i]=0;
	for (int i=1;i<=n+n;i++)
		rel[i]=low[i]=v[i]=0,b[i]=i,col[i]=0;
	w_time=0,top=0;
	bj=0;
}
void link(int x,int y)
{
	++ss,next[tail[x]]=ss,tail[x]=ss,sora[ss]=y,next[ss]=0;
}
void dfs(int x)
{
	w_time++;
	rel[x]=low[x]=w_time;
	st[++top]=x,v[x]=1;
	for (int i=x,ne;next[i];) {
		i=next[i],ne=sora[i];
		if (!v[ne]) {
			dfs(ne);
			low[x]=min(low[x],low[ne]);
		}
		else 
			if (v[ne]==1) low[x]=min(low[x],rel[ne]);
	}
	if (low[x]==rel[x]) {
		for (;st[top]!=x;top--) {
			int ne=st[top];
			b[ne]=x,v[ne]=2;
		} 
		top--;
		b[x]=x,v[x]=2;
		col[x]=++bj;
	}
}
int main()
{
	freopen("input.txt","r",stdin);
	for (;scanf("%d%d",&n,&m)==2;) {
		origin();
		int tot=0;
		for (int i=1;i<=n;i++) {
			scanf("%d",&age[i]);
			tot+=age[i];
		}
		for (int i=1;i<=n;i++)
			rank[i]=(age[i]*n>=tot);
		for (int i=1;i<=m;i++) {
			int x,y;
			scanf("%d%d",&x,&y);
			if (rank[x]!=rank[y]) {
				link(x+n,y);
				link(y+n,x);
			}
			else {
				link(x,y+n);
				link(y,x+n);
				link(x+n,y);
				link(y+n,x);
			}
		}
		for (int i=1;i<=n+n;i++)
			if (!v[i]) dfs(i);
		bool flag=1;
		for (int i=1;i<=n;i++)
			if (b[i]==b[i+n]) {
				flag=0;
				break;
			}
		if (!flag) {
			printf("No solution.\n");
			continue;
		}
		for (int i=1;i<=n;i++)
			if (col[b[i]]