负载是否平衡只与前两列有关，剩下的只要与前两列不重复就随便放。
第一列我们按1-n这样循环放，第二列每次找个数最少的那个服务器放。
#include#include #include using namespace std;const int MAXN = 110;int N, M;int mat[MAXN][MAXN];int cnt[MAXN];int ori[MAXN];bool vis[MAXN];void show(){ for ( int i = 1; i <= M; ++i ) { for ( int j = 1; j <= N; ++j ) { if ( j != 1 ) putchar(' '); printf( "%d", mat[i][j] ); } puts(""); } //puts("========="); return;}void FangHang( int *a ){ memset( vis, false, sizeof(vis) ); vis[ a[1] ] = true; vis[ a[2] ] = true; for ( int i = 3; i <= N; ++i ) { for ( int j = 1; j <= N; ++j ) if ( !vis[j] ) { a[i] = j; vis[j] = true; break; } } return;}int GetMin( int now ){ int minn = 1 << 30; int ansi; for ( int i = 1; i <= N; ++i ) { if ( i == now ) continue; if ( cnt[i] < minn ) { minn = cnt[i]; ansi = i; } } return ansi;}void solved(){ memset( cnt, 0, sizeof(cnt) ); for ( int i = 1; i <= M; ++i ) ++cnt[ mat[i][1] ]; int fang = 0; int cur = 1; for ( int i = 1; i <= N; ++i ) ori[i] = cnt[i]; while ( fang < M ) { int minn; int j = 1; while ( j <= M ) { minn = GetMin(cur); for ( ; j <= M; ++j ) { if ( mat[j][1] == cur && mat[j][2] == -1 ) { mat[j][2] = minn; ++cnt[ minn ]; ++fang; break; } } } for ( int i = 1; i <= N; ++i ) cnt[i] = ori[i]; ++cur; } for ( int i = 1; i <= M; ++i ) { FangHang( mat[i] ); } return;}int main(){ while ( scanf( "%d%d", &N, &M ) == 2 ) { memset( mat, -1, sizeof(mat) ); int i = 1, j = 1; while ( i <= M ) { mat[i][1] = j; ++i, ++j; if ( j > N ) j = 1; } solved(); show(); } return 0;}