CF1228D Complete Tripartite

Posted 2022-12-13 chy-2003

题目链接

问题分析

要求把点分成3组，每个组内没有边，每个点和每个不属于它这组的点之间都有边。

所以嘛，每组内的点连向的边都是相同的，和\(u\)不相邻的点都在\(u\)的同一组。

考虑到只有\(3\)组，所以直接\(O(n+m)\)暴力就好。可能需要通过代码理解一下

参考程序

#include <bits/stdc++.h>
using namespace std;
 
const int Maxn = 100010;
const int Maxm = 300010;
struct edge 
    int x, y;
    edge() 
    edge( int _x, int _y ) : x( _x ), y( _y ) 
    inline bool operator < ( const edge Other ) const 
        return x < Other.x || x == Other.x && y < Other.y;
    
    inline bool operator == ( const edge Other ) const 
        return x == Other.x && y == Other.y;
    
    inline bool operator > ( const edge Other ) const 
        return x > Other.x || x == Other.x && y > Other.y;
    
;
int n, m;
edge Edge[ Maxm << 1 ];
int Color[ Maxn ], App[ Maxn ], Cnt;
 
int main() 
    scanf( "%d%d", &n, &m );
    for( int i = 1; i <= m; ++i ) scanf( "%d%d", &Edge[ i ].x, &Edge[ i ].y );
    for( int i = 1; i <= m; ++i ) Edge[ i + m ] = edge( Edge[ i ].y, Edge[ i ].x );
    m <<= 1;
    sort( Edge + 1, Edge + m + 1 );
    for( int i = 1; i <= n; ++i ) 
        if( Color[ i ] ) continue;
        ++Cnt;
        if( Cnt > 3 ) 
            printf( "-1\n" );
            return 0;
        
 
        memset( App, 0, sizeof( App ) );
        int l = upper_bound( Edge + 1, Edge + m + 1, edge( i, 0 ) ) - Edge;
        int r = l - 1;
        while( Edge[ r + 1 ].x == i ) 
            ++r;
            App[ Edge[ r ].y ] = 1;
        
//      printf( "l = %d, r = %d\n", l, r );
//      printf( "A %d\n", i );
//      printf( "App : " ); for( int j = 1; j <= n; ++j ) printf( "%d ", App[ j ] ); printf( "\n" );
//      for( int j = l; j <= r; ++j ) printf( "%d %d\n", Edge[ j ].x, Edge[ j ].y );
 
        for( int j = 1; j <= n; ++j ) 
            if( App[ j ] == 0 && Color[ j ] != 0 ) 
                printf( "-1\n" );
                return 0;
            
            if( App[ j ] ) continue;
            Color[ j ] = Cnt;
            int L = upper_bound( Edge + 1, Edge + m + 1, edge( j, 0 ) ) - Edge;
            for( int k = l; k <= r; ++k ) 
                int R = L + ( k - l );
//              printf( "k = %d, R = %d, r = %d\n", k, R, l + ( k - l ) );
                if( Edge[ R ].x != j || Edge[ R ].y != Edge[ l + ( k - l ) ].y ) 
                    printf( "-1\n" );
                    return 0;
                
            
        
//      printf( "A %d\n", i );
//      for( int j = 1; j <= n; ++j ) printf( "%d ", Color[ j ] ); printf( "\n" );
    
    if( Cnt != 3 ) 
        printf( "-1\n" );
        return 0;
    
    for( int i = 1; i <= n; ++i ) printf( "%d ", Color[ i ] ); printf( "\n" );
    return 0;