CFR 596 C. Wilbur and Points ( Ad hoc )

Problem - C - Codeforces
The W array given in the input serves for constraining which points can be in those positions. For any ( x, y ), its special value is y - x, and it could only be in a position i, where W[ i ] == y - x. This means, we shall count and record the keys for each point and see if the W[] shows the exact spaces for them. For the same key, for example, ( x, y ) and ( x', y' ), where y - x == y' - x', and without losing generality x ≤ x', it is obvious that y ≤ y' also holds. So for each key we need a queue, storing positions for that key, and popping positions for the smallest ( x, y ) each time.

void solve(){
    int N; cin >> N;
    vp P( N );
    for( int i = 0; i < N; ++i )
        cin >> P[ i ].first >> P[ i ].second;
    vi W( N );
    map< int, queue< int > > svp_pos;
    map< int, int > cnt_spv;
    for( int i = 0; i < N; ++i )
        cin >> W[ i ],
        svp_pos[ W[ i ] ].push( i ),
        ++cnt_spv[ W[ i ] ];
    map< int, set< pii > > spv2p;
    for( int i = 0; i < N; ++i ){
        int key = P[ i ].second - P[ i ].first;
        spv2p[ key ].insert( P[ i ] );
        if( spv2p[ key ].size() > cnt_spv[ key ] ){
            cout << "NO\n";
            return;
        }
    }
    vp ans( N );
    for( auto t : spv2p ){
        for( pii p : t.second )
            ans[ svp_pos[ t.first ].front() ] = p,
            svp_pos[ t.first ].pop();
    }
    for( int i = 0; i + 1 < N; ++i ){
        if( ans[ i ].first >= ans[ i + 1 ].first and
            ans[ i ].second >= ans[ i + 1 ].second ){
            cout << "NO\n";
            return;
        }
    }
    cout << "YES\n";
    for( int i = 0; i < N; ++i )
        cout << ans[ i ].first << " " << ans[ i ].second << "\n";
}