3741. TJOI2014 拼图

题目大意

给定 n 个拼图碎片，现在要用这些碎片拼出一个 4×4 的正方形，要求每个碎片都被用到。若只有一组解，输出方案。

Data Constraint
n≤16

题解

考虑将问题转化成精确覆盖问题。
每个格子看成一列，因为每个碎片都要用到，所以每个碎片再开一列。
每个碎片可能匹配的位置都开一行，可以覆盖的位置对应的列就是矩阵中的1.

然后跑DLX即可。

SRC

#include
#include
#include
#include
#include
#include
using namespace std ;

#define N 10000 + 10

vector < int > Ans , Q ;

char ch[5][5] ;

int U[N] , D[N] , L[N] , R[N] , S[N] ;
int Row[N] , Col[N] , Belong[5][5] ;
int Cs[350] , Color[5][5] ;
int n , Cnt , Num ;

void AddRow( int r ) {
    int head = Cnt + 1 ;
    for (int i = 1 ; i <= Cs[0] ; i ++ ) {
        int now = Cs[i] ;
        ++ Cnt ;
        U[Cnt] = U[now] , D[Cnt] = now , L[Cnt] = Cnt - 1 , R[Cnt] = Cnt + 1 ;
        D[U[now]] = Cnt , U[now] = Cnt ;
        Row[Cnt] = r , Col[Cnt] = now ;
        S[now] ++ ;
    }
    L[head] = Cnt , R[Cnt] = head ;
}

void Remove( int c ) {
    R[L[c]] = R[c] ;
    L[R[c]] = L[c] ;
    for (int x = D[c] ; x != c ; x = D[x] ) {
        for (int y = R[x] ; y != x ; y = R[y] ) {
            D[U[y]] = D[y] , U[D[y]] = U[y] ;
            S[Col[y]] -- ;
        }
    }
}

void Restore( int c ) {
    for (int x = U[c] ; x != c ; x = U[x] ) {
        for (int y = L[x] ; y != x ; y = L[y] ) {
            S[Col[y]] ++ ;
            D[U[y]] = y , U[D[y]] = y ;
        }
    }
    R[L[c]] = c ;
    L[R[c]] = c ;
}

void DFS() {
    if ( R[0] == 0 ) {
        Num ++ ;
        if ( Num > 1 ) return ;
        Ans = Q ;
        return ;
    }
    int c = R[0] ;
    for (int x = R[0] ; x ; x = R[x] ) if ( S[x] < S[c] ) c = x ;
    Remove(c) ;
    for (int x = D[c] ; x != c ; x = D[x] ) {
        Q.push_back(x) ;
        if ( x == 222 ) { 
            x ++ ;
            x -- ;
        }
        for (int y = R[x] ; y != x ; y = R[y] ) Remove(Col[y]) ;
        DFS() ;
        Q.pop_back() ;
        if ( Num > 1 ) return ;
        for (int y = L[x] ; y != x ; y = L[y] ) Restore(Col[y]) ;
    }
    Restore(c) ;
}

void Print() {
    int size = Ans.size() ;
    for (int i = 0 ; i < size ; i ++ ) {
        int now = Ans[i] ;
        int w = now ;
        for (int x = R[now] ; x != now && Col[w] <= 16 ; x = R[x] ) {
            if ( Col[x] > 16 ) w = x ;
        }
        int type = Col[w] - 16 ;
        for (int x = R[w] ; x != w ; x = R[x] ) {
            int now = Col[x] ;
            int xx = now / 4 + (now % 4 != 0) ;
            int yy = now % 4 ? now % 4 : 4 ;
            Color[xx][yy] = type ;
        }
    }
    printf( "Yes, only one!\n" ) ;
    for (int i = 1 ; i <= 4 ; i ++ ) {
        for (int j = 1 ; j <= 4 ; j ++ ) printf( "%d" , Color[i][j] ) ;
        printf( "\n" ) ;
    }
}

int main() {
    freopen( "puzzle.in" , "r" , stdin ) ;
    freopen( "puzzle.out" , "w" , stdout ) ;
    while ( scanf( "%d" , &n ) != EOF ) {
        Ans.clear() ;
        memset( U , 0 , sizeof(U) ) ;
        memset( D , 0 , sizeof(D) ) ;
        memset( L , 0 , sizeof(L) ) ;
        memset( R , 0 , sizeof(R) ) ;
        memset( S , 0 , sizeof(S) ) ;
        memset( Row , 0 , sizeof(Row) ) ;
        memset( Col , 0 , sizeof(Col) ) ;
        for (int i = 0 ; i <= 16 + n ; i ++ ) {
            U[i] = D[i] = i ;
            L[i] = i - 1 , R[i] = i + 1 ;
        }
        L[0] = 16 + n , R[16+n] = 0 ;
        Cnt = 16 + n ;
        int tot = 0 ;
        for (int i = 1 ; i <= 4 ; i ++ )
            for (int j = 1 ; j <= 4 ; j ++ ) Belong[i][j] = ++ tot ;
        int Rowtot = 0 ;
        for (int i = 1 ; i <= n ; i ++ ) {
            int r , c ;
            scanf( "%d%d" , &r , &c ) ;
            for (int j = 1 ; j <= r ; j ++ ) {
                scanf( "\n" ) ;
                for (int k = 1 ; k <= c ; k ++ ) {
                    scanf( "%c" , &ch[j][k] ) ;
                }
            }
            for (int x = 1 ; x + r - 1 <= 4 ; x ++ ) {
                for (int y = 1 ; y + c - 1 <= 4 ; y ++ ) {
                    ++ Rowtot ;
                    Cs[0] = 0 ;
                    Cs[1] = 16 + i ;
                    for (int j = 1 ; j <= r ; j ++ ) {
                        for (int k = 1 ; k <= c ; k ++ ) {
                            if ( ch[j][k] == '0' ) continue ;
                            Cs[++Cs[0]] = Belong[x+j-1][y+k-1] ;
                        }
                    }
                    Cs[++Cs[0]] = 16 + i ;
                    AddRow( Rowtot ) ;
                }
            }
        }
        Num = 0 ;
        DFS() ;
        if ( Num == 0 ) { printf( "No solution\n" ) ; continue ; }
        if ( Num == 1 ) { Print() ; continue ; }
        if ( Num == 2 ) { printf( "Yes, many!\n" ) ; }
    }
    return 0 ;
}

以上.