poj 3207 2-sat

思路：把每一条线段当做点，则有“内”、“外”两种取值，然后在冲突的线段之间连边，判断是否有解即可。

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

const int N = 500;
const int M = 1000000;
int head[N * 2];
int s[N * 2];
bool mark[N * 2];
int n, e, c;

struct Edge
{
    int v, next;
} edge[M];

void addEdge( int u, int v )
{
    edge[e].v = v;
    edge[e].next = head[u];
    head[u] = e++;
}

void init()
{
    e = 0;
    memset( head, -1, sizeof(head) );
    memset( mark, false, sizeof(mark) );
}

bool dfs( int u )
{
    if ( mark[u ^ 1] ) return false;
    if ( mark[u] ) return true;
    mark[u] = true;
    s[c++] = u;
    for ( int i = head[u]; i != -1; i = edge[i].next )
    {
        int v = edge[i].v;
        if ( !dfs(v) ) return false;
    }
    return true;
}

bool solve()
{
    for ( int i = 0; i < 2 * n; i += 2 )
    {
        if ( !mark[i] && !mark[i + 1] )
        {
            c = 0;
            if ( !dfs(i) )
            {
                while ( c )
                {
                    c--;
                    mark[s[c]] = false;
                }
                if ( !dfs( i + 1 ) ) return false;
            }
        }
    }
    return true;
}

struct T 
{
    int x, y;
} t[N];

int main ()
{
    scanf("%d%d", &n, &n);
    init();
    for ( int i = 0; i < n; i++ )
    {
        scanf("%d%d", &t[i].x, &t[i].y);
        if ( t[i].x > t[i].y ) swap( t[i].x, t[i].y );
        for ( int j = 0; j < i; j++ )
        {
            if ( ( t[i].x < t[j].x && t[i].y < t[j].y && t[i].y > t[j].x ) 
              || ( t[i].x > t[j].x && t[i].y > t[j].y && t[i].x < t[j].y ) )
            {
                int li = i << 1, ri = li | 1;
                int lj = j << 1, rj = lj | 1;
                addEdge( li, rj );
                addEdge( ri, lj );
                addEdge( lj, ri );
                addEdge( rj, li );
            }
        }
    }
    if ( solve() )
    {
        puts("panda is telling the truth...");
    }
    else
    {
        puts("the evil panda is lying again");
    }
    return 0;
}