POJ 3207 Ikki's Story IV - Panda's Trick(2-SAT)
POJ 3207 Ikki's Story IV - Panda's Trick(2-SAT)
Description
liympanda, one of Ikki’s friend, likes playing games with Ikki. Today after minesweeping with Ikki and winning so many times, he is tired of such easy games and wants to play another game with Ikki.
liympanda has a magic circle and he puts it on a plane, there are n points on its boundary in circular border: 0, 1, 2, …, n − 1. Evil panda claims that he is connecting m pairs of points. To connect two points, liympanda either places the link entirely inside the circle or entirely outside the circle. Now liympanda tells Ikki no two links touch inside/outside the circle, except on the boundary. He wants Ikki to figure out whether this is possible…
Despaired at the minesweeping game just played, Ikki is totally at a loss, so he decides to write a program to help him.
Input
The input contains exactly one test case.
In the test case there will be a line consisting of of two integers: n and m (n ≤ 1,000, m ≤ 500). The following m lines each contain two integers ai and bi, which denote the endpoints of the ith wire. Every point will have at most one link.
Output
Output a line, either “panda is telling the truth...” or “the evil panda is lying again”.
Sample Input
4 2
0 1
3 2Sample Output
panda is telling the truth...题意：有一个圆，给出一些边连接着两个点，边可以从圆里连，也可以从圆外连，问是否可以不相交
对于边i,j限制条件为不相交，即不在同一个集合中
因此我们将这个问题转化为了2-SAT问题
设i表示边i在圆内，i'表示i在圆外
若(i,j)在圆内相交，那么它们在圆外也一定相交
如果边i,j在圆内相交
那么就从i连向j'（i内j外），从j'连向i（i内j外），从j连向i'（j内i外），从i'连向j（j内i外）
然后来一遍tarjan就好了
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stack>
#define Pair pair<int,int>
#define F first
#define S second
using namespace std;
const int MAXN=1e6+10;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<20,stdin),p1==p2)?EOF:*p1++)
char buf[1<<20],*p1=buf,*p2=buf;
inline int read()
{ char c=getchar();int x=0,f=1;
while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
return x*f;
}
Pair p[MAXN];
struct node
{
int u,v,nxt;
}edge[MAXN];
int head[MAXN],num=1;
inline void AddEdge(int x,int y)
{
edge[num].u=x;
edge[num].v=y;
edge[num].nxt=head[x];
head[x]=num++;
}
int dfn[MAXN],low[MAXN],vis[MAXN],color[MAXN],colornum=0,tot=0;
stack<int>s;
void tarjan(int now)
{
dfn[now]=low[now]=++tot;
s.push(now);
vis[now]=1;
for(int i=head[now];i!=-1;i=edge[i].nxt)
{
if(!dfn[edge[i].v])
tarjan(edge[i].v),low[now]=min(low[now],low[edge[i].v]);
if(vis[edge[i].v]) low[now]=min(low[now],dfn[edge[i].v]);
}
if(dfn[now]==low[now])
{
colornum++;
int h=0;
do
{
h=s.top();s.pop();
vis[h]=0;
color[h]=colornum;
}while(h!=now);
}
}
int main()
{
#ifdef WIN32
freopen("a.in","r",stdin);
#else
#endif
memset(head,-1,sizeof(head));
int N=read(),M=read();
for(int i=1;i<=M;i++)
{
p[i].F=read(),p[i].S=read();
if(p[i].F>p[i].S) swap(p[i].F,p[i].S);
}
for(int i=1;i<=M;i++)
{
for(int j=i+1;j<=M;j++)
{
if((p[j].F>=p[i].F&&p[j].F<=p[i].S&&p[j].S>=p[i].S)||
(p[j].F<=p[i].F&&p[j].S>=p[i].F&&p[j].S<=p[i].S))
AddEdge(i,j+M),
AddEdge(j,i+M),
AddEdge(j+M,i),
AddEdge(i+M,j);
}
}
for(int i=1;i<=M;i++)
if(!dfn[i])
tarjan(i);
bool flag=1;
for(int i=1;i<=M;i++)
if(color[i]==color[i+M])
{printf("the evil panda is lying again\n");flag=0;break;}
if(flag==1) printf("panda is telling the truth...\n");
return 0;
}