【代码】POJ 2762

// 题目来源：POJ 2762
// 题目大意：给定一个有向图，问该图是否弱连通（若图中任意两点之间可以到达，即可以i到j或者j到i，该图即为弱连通）
// 解决方法：强连通分量缩点，当且仅当该图的拓扑序列唯一的时候才成立，那么只需记录当前入度为0的个数，当多于1个时不成立

#include <cstdio>
#include <string>
#define _ 60002
#define o 10002
using namespace std;

void link( int, int );
void link2( int, int );
void tarjan( int );
bool judge( );
void topsort( int );

int dfn[ o ], low[ o ], h[ o ], h2[ o ], id[ o ], od[ o ], stack[ o ], code[ o ];
int next[ _ ], g[ _ ], next2[ _ ], g2[ _ ];
int q, n, m, t, t2, top, cnt, index, sid;
bool ins[ o ];
bool flag;

int main( )
{
freopen( "2762.in", "r", stdin );
freopen( "2762.out", "w", stdout );
scanf( "%d", &q );
int aa, bb;
while( q-- )
{
memset( dfn, 0, sizeof( dfn ) );
memset( low, 0, sizeof( low ) );
memset( next, 0, sizeof( next ) );
memset( h, 0, sizeof( h ) );
memset( next2, 0, sizeof( next2 ) );
memset( h2, 0, sizeof( h2 ) );
memset( id, 0, sizeof( id ) );
memset( od, 0, sizeof( od ) );
memset( code, 0, sizeof( code ) );
t = 0;
t2 = 0;
cnt = 0;
scanf( "%d%d", &n, &m );
for( int i = 1; i <= m; i++ )
{
scanf( "%d%d", &aa, &bb );
link( aa, bb );
}
for( int i = 1; i <= n; i++ )
if( !dfn[ i ] ) tarjan( i );
if( judge( ) )
printf( "Yes\n" );
else
printf( "No\n" );
}
return 0;
}

void link( int aa, int bb )
{
next[ ++t ] = h[ aa ];
h[ aa ] = t;
g[ t ] = bb;
}

void link2( int aa, int bb )
{
next2[ ++t2 ] = h2[ aa ];
h2[ aa ] = t2;
g2[ t2 ] = bb;
}

void tarjan( int i )
{
int j;
dfn[ i ] = low[ i ] = ++index;
stack[ ++top ] = i;
ins[ i ] = true;
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 ] && dfn[ j ] < low[ i ] )
low[ i ] = dfn[ j ];
}
if( low[ i ] == dfn[ i ] )
{
cnt++;
do
{
j = stack[ top-- ];
code[ j ] = cnt;
ins[ j ] = false;
}
while( i != j );
}
}

bool judge( )
{
int j;
flag = true;
sid = 0;
for( int i = 1; i <= n; i++ )
for( int k = h[ i ]; k; k = next[ k ] )
{
j = g[ k ];
if( code[ i ] != code[ j ] )
{
id[ code[ j ] ]++;
link2( code[ i ], code[ j ] );
}
}
for( int i = 1; i <= cnt; i++ )
if( id[ i ] == 0 ) sid++;
if( sid > 1 ) return 0;
for( int i = 1; i <= cnt; i++ )
if( id[ i ] == 0 ) topsort( i );
if( flag ) return 1;
return 0;
}

void topsort( int i )
{
int j;
sid--;
id[ i ] = -1;
for( int k = h2[ i ]; k; k = next2[ k ] )
{
j = g2[ k ];
id[ j ]--;
if( id[ j ] == 0 ) sid++;
}
if( sid > 1 ) flag = false;
for( int k = h2[ i ]; k; k = next2[ k ] )
{
j = g2[ k ];
if( id[ j ] == 0 ) topsort( j );
}
}