poj3683(2SAT)

解题思路：大概讲一下建图思路：首先将一个婚礼分成有个点，分别表示在开头神父到，和在末尾神父到。然后判断各个婚礼之间的矛盾情况，按2SAT传统连边。

#include<cstdio>
#include<cstring>
#include<stack>
#include<algorithm>
using namespace std;
const int MAXN=10080;
struct node
{
	int st,ed;
}t[2008];
struct edge
{
	int u,v,next;
}s[4000008];
int cnt,head[2008],dfn[2008],low[2008],ins[2008],top,color[2008],num,tim,stck[2008],oppo[2008];
int deg[2008],jd[2008],n;
stack<int> q;
bool judge(int i1,int i2)
{
	if(t[i1].ed<=t[i2].st||t[i2].ed<=t[i1].st)
	return false;
	return true;
}
void add_edge(int u,int v)
{
	s[cnt].u=u;
	s[cnt].v=v;
	s[cnt].next=head[u];
	head[u]=cnt++;
}
void tarjan(int u)
{
	int tp;
	//printf("%d\n",u);
	ins[u]=1;
	stck[top++]=u;
	dfn[u]=low[u]=tim++;
	for(int i=head[u];i!=-1;i=s[i].next)
	{
		int v=s[i].v;
		if(!dfn[v])
		{
			tarjan(v);
			low[u]=min(low[u],low[v]);
		}else
		if(ins[v])
		{
			low[u]=min(low[u],dfn[v]);
		}
	}
	if(low[u]==dfn[u])
	{
		num++;
		do
		{
			tp=stck[--top];
			ins[tp]=0;
			color[tp]=num;
		}while(tp!=u);
	}
}
void topo()
{
	while(!q.empty()) q.pop();
	for(int i=1;i<=num;i++)
	{
		if(deg[i]==0) q.push(i);
	}
	while(!q.empty())
	{
		int u=q.top();
		q.pop();
		if(jd[u]==0) jd[u]=1,jd[oppo[u]]=-1;
		for(int i=head[u];i!=-1;i=s[i].next)
		{
			int v=s[i].v;
			deg[v]--;
			if(deg[v]==0)
			q.push(v);
		}
	}
}
void solve()
{
	memset(dfn,0,sizeof(dfn));
	memset(ins,0,sizeof(ins));
	top=0;
	num=0;
	tim=1;
	for(int i=0;i<2*n;i++)
	{
		if(!dfn[i])
		tarjan(i);
	}
	//printf("%d",num);
	for(int i=0;i<n;i++)
	{
		if(color[i]==color[i+n])
		{
			printf("NO\n");
			return;
		}
		oppo[color[i]]=color[i+n];
		oppo[color[i+n]]=color[i];
	}
	int tp=cnt;
	memset(head,-1,sizeof(head));
	memset(deg,0,sizeof(deg));
	for(int i=0;i<tp;i++)
	{
		if(color[s[i].v]==color[s[i].u]) continue;
		add_edge(color[s[i].v],color[s[i].u]);
		deg[color[s[i].u]]++;
	}
	memset(jd,0,sizeof(jd));
	topo();
	printf("YES\n");
	for(int i=0;i<n;i++)
	{
		//printf("%d*",jd[color[i]]);
		if(jd[color[i]]==1)
		{
			//printf("*%d*",i);
			printf("%02d:%02d %02d:%02d\n",t[i].st/60,t[i].st%60,t[i].ed/60,t[i].ed%60);
		}else
		{
			printf("%02d:%02d %02d:%02d\n",t[i+n].st/60,t[i+n].st%60,t[i+n].ed/60,t[i+n].ed%60);
		}
	}
}
int main()
{
	int a1,a2,b1,b2,c;
//	freopen("t.txt","r",stdin);
	while(scanf("%d",&n)!=EOF)
	{
		for(int i=0;i<n;i++)
		{
			scanf("%d:%d %d:%d %d",&a1,&a2,&b1,&b2,&c);
			//printf("%d:%d %d:%d %d\n",a1,a2,b1,b2,c);
			t[i].st=a1*60+a2;
			t[i+n].ed=b1*60+b2;
			t[i].ed=t[i].st+c;
			t[i+n].st=t[i+n].ed-c;
		}
		memset(head,-1,sizeof(head));
		cnt=0;
		for(int i=0;i<n;i++)
		{
			for(int j=i+1;j<n;j++)
			{
				if(judge(i,j))
				{
					//printf("%d %d\n",i,j);
					add_edge(i,j+n);
					add_edge(j,i+n);
				}
				if(judge(i+n,j))
				{
					//printf("%d %d\n",i+n,j);
					add_edge(i+n,j+n);
					add_edge(j,i);
				}
				if(judge(i+n,j+n))
				{
					//printf("%d %d\n",i+n,j+n);
					add_edge(i+n,j);
					add_edge(j+n,i);
				}
				if(judge(i,j+n))
				{
					//printf("%d %d\n",i,j+n);
					add_edge(i,j);
					add_edge(j+n,i+n);
				}				
			}
		}
		solve();
	}
	return 0;
 }