POJ 3207 Ikki's Story IV

2018-03-02 08:28:44来源:cnblogs.com作者:Panda's Trick(2-SAT) - 自为风人点击

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Time Limit: 1000MS Memory Limit: 131072K
Total Submissions: 10870 Accepted: 3988

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 20 13 2

Sample 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 secondusing 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;    }

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