二分图匹配,每个点只判断与其周围各点的匹配情况,难点在于对于一个编号求期所在的行列,以及对于给出的坐标求编号。
#include < iostream >
#include < cstdio >
#include < cstdlib >
#include < cstring >
using namespace std;
#define maxn 70
int n, m, h;
bool map[maxn][maxn];
int xM[maxn * maxn], yM[maxn * maxn];
bool chk[maxn * maxn];
int uN, vN;
int dir[ 4 ][ 2 ] =
{
{ 1 , 0 },
{ 0 , 1 },
{ - 1 , 0 },
{ 0 , - 1 } };
int getx( int a)
{
if (m & 1 )
return (a * 2 + 1 ) / m;
return a / (m / 2 );
}
int gety( int a)
{
if (m & 1 )
return (a * 2 + 1 ) % m;
int x = getx(a);
return 1 - x % 2 + (a * 2 - x * m);
}
int hash( int x, int y)
{
if (x < 0 || y < 0 || x >= n || y >= m)
return - 1 ;
if (map[x][y])
return - 1 ;
if (m & 1 )
return (x * m + y) / 2 ;
return x * (m / 2 ) + y / 2 ;
}
bool SearchPath( int u)
{
int x = getx(u);
int y = gety(u);
if (map[x][y])
return false ;
for ( int i = 0 ; i < 4 ; i ++ )
{
int v = hash(x + dir[i][ 0 ], y + dir[i][ 1 ]);
if (v != - 1 && ! chk[v])
{
chk[v] = true ;
if (yM[v] == - 1 || SearchPath(yM[v]))
{
yM[v] = u;
xM[u] = v;
return true ;
}
}
}
return false ;
}
int MaxMatch()
{
int u, ret = 0 ;
memset(xM, - 1 , sizeof (xM));
memset(yM, - 1 , sizeof (yM));
for (u = 0 ; u < uN; u ++ )
if (xM[u] == - 1 )
{
memset(chk, 0 , sizeof (chk));
if (SearchPath(u))
ret ++ ;
}
return ret;
}
int main()
{
// freopen("t.txt", "r", stdin);
scanf( " %d%d%d " , & n, & m, & h);
memset(map, 0 , sizeof (map));
uN = m * n / 2 ;
vN = m * n - uN;
for ( int i = 0 ; i < h; i ++ )
{
int a, b;
scanf( " %d%d " , & a, & b);
a -- ;
b -- ;
map[b][a] = true ;
}
if ((m * n - h) & 1 )
{
printf( " NO\n " );
return 0 ;
}
int ans = MaxMatch();
if (ans < (m * n - h) / 2 )
printf( " NO\n " );
else
printf( " YES\n " );
return 0 ;
}