POJ 3207 Ikki＆＃39; s Story IV – Panda＆＃39; s Trick (2-SAT)

Hello everyone , I meet you again , I'm the king of the whole stack .

Position address ： Find contradictions . The contradiction is (2,5) and (3,6) Only one of these two can be outside , One is inside . Use this contradictory relationship to build a map .

It can be used outside and inside as 1 and 0, Finally, infer whether each pair has contradictions .

The code is as follows ：

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <stdlib.h>
#include <math.h>
#include <ctype.h>
#include <queue>
#include <map>
#include <set>
#include <algorithm>

using namespace std;
#define LL __int64
const int INF=0x3f3f3f3f;
int head[1100], cnt, top, ans, index;
int dfn[1100], low[1100], instack[1100], stak[1100], belong[1100];
struct node
{
    int u, v, next;
}edge[1000000];
struct N
{
    int l, r;
}xian[10000];
void add(int u, int v)
{
    edge[cnt].v=v;
    edge[cnt].next=head[u];
    head[u]=cnt++;
}
int pan(N x, N y)
{
    if((x.l>y.l&&x.l<y.r&&x.r>y.r)||(x.r>y.l&&x.r<y.r&&x.l<y.l))
        return 1;
    return 0;
}
void tarjan(int u)
{
    dfn[u]=low[u]=++index;
    instack[u]=1;
    stak[++top]=u;
    for(int i=head[u];i!=-1;i=edge[i].next)
    {
        int v=edge[i].v;
        if(!dfn[v])
        {
            tarjan(v);
            low[u]=min(low[u],low[v]);
        }
        else if(instack[v])
        {
            low[u]=min(dfn[v],low[u]);
        }
    }
    if(dfn[u]==low[u])
    {
        ans++;
        while(1)
        {
            int v=stak[top--];
            instack[v]=0;
            belong[v]=ans;
            if(u==v) break;
        }
    }
}
void init()
{
    memset(head,-1,sizeof(head));
    cnt=top=ans=index=0;
    memset(dfn,0,sizeof(dfn));
    memset(instack,0,sizeof(instack));
}
int main()
{
    int n, m, i, j;
    scanf("%d%d",&n,&m);
    init();
    for(i=0;i<m;i++)
    {
        scanf("%d%d",&xian[i].l,&xian[i].r);
        if(xian[i].l>xian[i].r) swap(xian[i].l,xian[i].r);
    }
    for(i=0;i<m;i++)
    {
        for(j=0;j<i;j++)
        {
            if(pan(xian[i],xian[j]))
            {
                add(i<<1,j<<1|1);
                add(j<<1,i<<1|1);
                add(i<<1|1,j<<1);
                add(j<<1|1,i<<1);
                //printf("%d %d\n%d %d\n",i<<1,j<<1|1,j<<1,i<<1|1);
            }
        }
    }
    for(i=0;i<2*m;i++)
    {
        if(!dfn[i])
            tarjan(i);
    }
    int flag=0;
    for(i=0;i<m;i++)
    {
        if(belong[i<<1]==belong[i<<1|1])
        {
            flag=1;
            //printf("%d\n",i);
            break;
        }
    }
    if(flag) puts("the evil panda is lying again");
    else
        puts("panda is telling the truth...");
    return 0;
}

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