【代码】POJ 3207

// 题目来源：POJ 3207// 题目大意：圆上顺序排列n个点，现要在一些点间连边，规定边只能在圆内或圆外，求有没有可能不相交// 解决方法：连边只能在里面或者外面，这就是2-sat问题，直接缩点判断即可 #include <cstdio>#define o 5002void link( int, int );void tarjan( int );int next[ o ], h[ o ], g[ o ], a[ o ], b[ o ], code[ o ], dfn[ o ], low[ o ], stack[ o ];int n, m, t, top, index, cnt, i = 0, j;bool ins[ o ];int main( ){    freopen( "3207.in", "r", stdin );    freopen( "3207.out", "w", stdout );    scanf( "%d%d", &n, &m );    for( int i = 0; i <= m-1; i++ )    {        scanf( "%d%d", &a[ i ], &b[ i ] );        if( a[ i ] > b[ i ] )        {            int temp = a[ i ];            a[ i ] = b[ i ];            b[ i ] = temp;        }    }    for( int i = 0; i <= m-1; i++ )        for( int j = i + 1; j <= m-1; j++ )            if( a[ i ] < a[ j ] && b[ i ] > a[ j ] && b[ i ] < b[ j ] || b[ i ] > b[ j ] && a[ i ] > a[ j ] && a[ i ] < b[ j ] )            {                link( i*2, j*2+1 );                link( i*2+1, j*2 );                link( j*2, i*2+1 );                link( j*2+1, i*2 );            }    for( int i = 0; i <= 2*m-1; i++ )        if( !dfn[ i ] ) tarjan( i );    for( int i = 0; i <= 2*m-1; i++ )        if( code[ i ] == code[ i^1 ] )        {            printf( "the evil panda is lying again" );            return 0;        }    printf( "panda is telling the truth..." );    return 0;}void link( int aa, int bb ){    next[ ++t ] = h[ aa ];    h[ aa ] = t;    g[ t ] = bb;}void tarjan( int i ){    int j;    dfn[ i ] = low[ i ] = ++index;    stack[ ++top ] = i;    ins[ i ] = 1;    for( int k = h[ i ]; k; k = next[ k ] )    {        j = g[ k ];        if( !dfn[ j ] )    {            tarjan( j );            if( low[ j ] < low[ i ] ) low[ i ] = low[ j ];        }        else if( ins[ j ] )            if( dfn[ j ] < low[ i ] ) low[ i ] = dfn[ j ];    }    if( low[ i ] == dfn[ i ] )    {        cnt++;        do        {            j = stack[ top-- ];            code[ j ] = cnt;            ins[ j ] = 0;        }        while( i != j );    }}