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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#285836#7607. The Doubling Game 2SortingWA 2ms25144kbC++205.8kb2023-12-16 23:26:312023-12-16 23:26:32

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你现在查看的是最新测评结果

  • [2023-12-16 23:26:32]
  • 评测
  • 测评结果:WA
  • 用时:2ms
  • 内存:25144kb
  • [2023-12-16 23:26:31]
  • 提交

answer

#include<bits/stdc++.h>
using namespace std ;
typedef long long ll ;
typedef unsigned long long ull ;
typedef pair < int , int > pii ;
typedef vector < int > vi ;
#define fi first
#define se second
mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());

#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()


const int MAXN = 3e5 + 7 ;
const int MOD = 1e9 + 7 ;
const int LOG = 20 ;

int n ;
vector < int > v[ MAXN ] ;
int subtree[ MAXN ] ;

bool cmp ( int x , int y ) {
    return ( subtree[ x ] < subtree[ y ] ) ;
}


ll dp[ MAXN ][ LOG ] ; //
ll root[ MAXN ][ LOG ] ; //
ll complete[ MAXN ] ;

int mxpw[ MAXN ] ;
ll f[ ( 1 << LOG ) ][ LOG ] ;
int bit_rv[ 2 * ( 1 << LOG ) ] ;

void dfs ( int x , int prv ) {

    vector < int > children ;
    for ( auto y : v[ x ] ) {
        if ( y == prv ) { continue ; }
        dfs ( y , x ) ;
        children.push_back ( y ) ;
    }
    sort ( children.begin ( ) , children.end ( ) , cmp ) ;
    mxpw[ x ] = 0 ;
    subtree[ x ] = 1 ;
    for ( int i = 0 ; i < LOG ; ++ i ) {
        f[ 0 ][ i ] = 0 ;
    }
    f[ 0 ][ LOG - 1 ] = 1 ;
    for ( auto y : children ) {
        subtree[ x ] += subtree[ y ] ;
        int rem = mxpw[ x ] ;
        mxpw[ x ] = max ( mxpw[ x ] , min ( mxpw[ x ] + 1 , mxpw[ y ] + 1 ) ) ;
        if ( rem < mxpw[ x ] ) {
            int lo = ( 1 << rem ) ;
            int hi = 2 * lo - 1 ;
            for ( int i = lo ; i <= hi ; ++ i ) {
                for ( int j = 0 ; j < LOG ; ++ j ) {
                    f[ i ][ j ] = 0 ;
                }
            }
        }
        /**
        if ( x == 1 ) {
            printf ( "--- child %d\n" , y ) ;
        }
        **/
        for ( int i = ( 1 << mxpw[ x ] ) - 1 ; i >= 0 ; -- i ) {
            for ( int j = 0 ; j <= mxpw[ y ] ; ++ j ) { 
                if ( ( i & ( 1 << j ) ) == 0 ) { 
                    for ( int dig = 0 ; dig <= mxpw[ x ] ; ++ dig ) {
                        if ( j == dig ) { continue ; }
                        f[ ( i + ( 1 << j ) ) ][ dig ] = ( f[ ( i + ( 1 << j ) ) ][ dig ] + f[ i ][ dig ] * root[ y ][ j ] ) % MOD ;
                        /**
                        if ( x == 1 && i + ( 1 << j ) == 1 && dig == 1 ) {
                            printf ( "add %lld == %lld * %lld at %d %d %d\n" , f[ i ][ dig ] * root[ y ][ j ] , f[ i ][ dig ] , root[ y ][ j ] , i , j , dig ) ;
                        }
                        **/
                    }
                    f[ ( i + ( 1 << j ) ) ][ LOG - 1 ] = ( f[ ( i + ( 1 << j ) ) ][ LOG - 1 ] + f[ i ][ LOG - 1 ] * root[ y ][ j ] ) % MOD ;
                }
                /**
                if ( x == 1 && i == 1 && j == 1 ) {
                    printf ( "add nw %lld idle %lld \n" , f[ i ][ LOG - 1 ] * dp[ y ][ j ] , f[ i ][ j ] * complete[ y ] ) % MOD ; 
                }
                if ( x == 1 && i == 0 && j == 1 ) { 
                    printf ( "add to 0 1 nw %lld idle %lld \n" , f[ i ][ LOG - 1 ] * dp[ y ][ j ] , f[ i ][ j ] * complete[ y ] ) % MOD ; 
                }
                **/
                f[ i ][ j ] = ( f[ i ][ LOG - 1 ] * dp[ y ][ j ] + f[ i ][ j ] * complete[ y ] ) % MOD ;
            }
            f[ i ][ LOG - 1 ] = ( f[ i ][ LOG - 1 ] * complete[ y ] ) % MOD ;
        }
    }
    /**
    for ( int i = 0 ; i < ( 1 << mxpw[ x ] ) ; ++ i ) {
        for ( int j = 0 ; j <= mxpw[ x ] ; ++ j )  {
            printf ( "vertex %d --> f[ %d ][ %d ] = %lld\n" , x , i , j , f[ i ][ j ] ) ;
        }
        printf ( "vertex %d --> f[ %d ][ -1 ] = %lld\n" , x , i , f[ i ][ LOG - 1 ] ) ;
    }
    **/
    for ( int mask = 0 ; mask < ( 1 << mxpw[ x ] ) ; ++ mask ) {
        int lw = bit_rv[ ( ( ~mask ) & -( ~mask ) ) ] ; // smallest 0 bit id
        for ( int p1 = 0 ; p1 < LOG ; ++ p1 ) {
            if ( ( mask & ( 1 << p1 ) ) > 0 ) { continue ; }
            if ( p1 == LOG - 1 ) { // 
                if ( ( ( mask + ( 1 << lw ) ) & ( mask + ( 1 << lw ) + 1 ) ) == 0 ) { 
                    dp[ x ][ lw ] = ( dp[ x ][ lw ] + f[ mask ][ p1 ] ) % MOD ;
                }
                if ( ( mask & ( mask + 1 ) ) == 0 ) {
                    root[ x ][ lw ] = ( root[ x ][ lw ] + f[ mask ][ p1 ] ) % MOD ;
                    /**
                    if ( x == 1 ) {
                        printf ( "-- add %d %d %lld\n" , mask , p1 , f[ mask ][ p1 ] ) ;
                    }
                    **/
                    complete[ x ] = ( complete[ x ] + f[ mask ][ p1 ] ) % MOD ;
                }
            }
            else {
                if ( ( mask + ( 1 << lw ) ) == ( 1 << p1 ) - 1 ) {
                    dp[ x ][ lw ] = ( dp[ x ][ lw ] + f[ mask ][ p1 ] ) % MOD ;
                }
                if ( mask == ( 1 << p1 ) - 1 ) {
                    /**
                    if ( x == 1 ) {
                        printf ( "-- add %d %d %lld\n" , mask , p1 , f[ mask ][ p1 ] ) ;
                    }
                    **/
                    complete[ x ] = ( complete[ x ] + f[ mask ][ p1 ] ) % MOD ;
                }
            }
        }
    }
    /**
    printf ( "complete[ %d ] = %lld\n" , x , complete[ x ] ) ;
    for ( int i = 0 ; i < ( 1 << mxpw[ x ] ) ; ++ i ) {
        printf ( "dp[ %d ][ %d ] = %lld\n" , x , i , dp[ x ][ i ] ) ;
        printf ( "root[ %d ][ %d ] = %lld\n" , x , i , root[ x ][ i ] ) ;
    }
    **/
}

void solve ( ) {
    cin >> n ;
    for ( int i = 1 , x , y ; i < n ; ++ i ) {
        cin >> x >> y ;
        v[ x ].push_back ( y ) ;
        v[ y ].push_back ( x ) ;
    }
    for ( int i = 0 ; i < 2 * ( 1 << LOG ) ; ++ i ) {
        bit_rv[ i ] = -1 ;
    }
    for ( int i = 0 ; i <= LOG ; ++ i ) {
        bit_rv[ ( 1 << i ) ] = i ;
    }
    dfs ( 1 , -1 ) ;
    cout << complete[ 1 ] << "\n" ;
}

int main ( ) {
    ios_base :: sync_with_stdio ( false ) ;
    cin.tie ( NULL ) ;
    int t = 1 ; // cin >> t ;
    while ( t -- ) { solve ( ) ; }
    return 0 ;
}


详细

Test #1:

score: 100
Accepted
time: 2ms
memory: 25076kb

input:

5
1 2
1 3
1 4
4 5

output:

21

result:

ok single line: '21'

Test #2:

score: 0
Accepted
time: 0ms
memory: 25144kb

input:

1

output:

1

result:

ok single line: '1'

Test #3:

score: -100
Wrong Answer
time: 0ms
memory: 25144kb

input:

128
11 32
116 81
65 4
117 47
5 81
104 30
61 8
82 59
95 20
92 29
29 127
97 39
123 33
59 128
115 33
83 67
74 16
77 33
64 73
124 123
8 127
61 51
101 122
35 90
119 116
112 27
81 93
109 123
54 1
119 100
116 16
65 47
67 27
22 105
76 87
36 39
27 96
72 31
91 123
21 105
118 12
110 48
121 72
14 115
24 16
106 ...

output:

541227059

result:

wrong answer 1st lines differ - expected: '508800953', found: '541227059'