HDU 3207 Ikki's Story IV-panda's trick (graph theory-2sat, graph theory-Tarjan)

Ikki's Story IV-panda's trick

Time limit:1000 ms		Memory limit:131072 K
Total submissions:7821		Accepted:2892

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 areNPoints on its boundary in circular border: 0, 1, 2 ,...,N? 1. edevil panda claims that he is ing 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, cycle T 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 two integers:NAndM(N≤ 1,000,M≤ 500). The followingMLines each contain two integersAIAndBi, Which denote the endpoints ofITh 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...

Source

Poj monthly -- 2007.03.04, Ikki

Question:

N points, M line segments, and line segments can be placed outside and inside the ring and asked if there is a solution that does not overlap.

Solution:

The 2sat method is used to determine whether a conflicting edge exists after Tarjan is finished based on the link edge of the conflicting link.

Solution code:

#include <iostream>#include <cstdio>#include <vector>#include <algorithm>using namespace std;const int maxm=510000;const int maxn=1100;struct edge{    int u,v,next;    edge(int u0=0,int v0=0){        u=u0;v=v0;    }}e[maxm];int n,m,head[maxn],color[maxn],dfn[maxn],low[maxn],cnt,nc,index;bool mark[maxn];vector < pair<int,int> > v;vector <int> vec;void addedge(int u,int v){    e[cnt]=edge(u,v);e[cnt].next=head[u];head[u]=cnt++;}void input(){    vec.clear();    cnt=nc=index=0;    for(int i=0;i<=2*m;i++){        color[i]=head[i]=-1;        mark[i]=false;        dfn[i]=0;    }    v.clear();    v.resize(m);    for(int i=0;i<m;i++){        scanf("%d%d",&v[i].first,&v[i].second);        if(v[i].first>v[i].second) swap(v[i].first,v[i].second);    }}void tarjan(int s){    dfn[s]=low[s]=++index;    mark[s]=true;    vec.push_back(s);    for(int i=head[s];i!=-1;i=e[i].next){        int d=e[i].v;        if(!dfn[d]){            tarjan(d);            low[s]=min(low[d],low[s]);        }else if(mark[d]){            low[s]=min(low[s],dfn[d]);        }    }    if(dfn[s]==low[s]){        nc++;        int d;        do{            d=vec.back();            vec.pop_back();            color[d]=nc;            mark[d]=false;        }while(d!=s);    }}void solve(){    for(int i=0;i<m;i++){        for(int j=i+1;j<m;j++){            if( v[i].first>=v[j].second || v[j].first>=v[i].second ) continue;            if( ( v[i].first<=v[j].first && v[j].second<=v[i].second ) || ( v[j].first<=v[i].first && v[i].second<=v[j].second ) ) continue;            addedge(i,j+m);            addedge(i+m,j);            addedge(j,i+m);            addedge(j+m,i);        }    }    for(int i=0;i<2*m;i++){        if(!dfn[i]) tarjan(i);    }    for(int i=0;i<m;i++){        if(color[i]==color[i+m]){            printf("the evil panda is lying again\n");            return;        }    }    printf("panda is telling the truth...\n");}int main(){    while(scanf("%d%d",&n,&m)!=EOF){        input();        solve();    }    return 0;}