sgu 433 Japhshan and Ramshut

http://acm.sgu.ru/problem.php?contest=0&problem=433

题目大意：要求使用一个长为L，宽为1的矩形，刚好填充一个大的矩形。
解法：比较裸的DLX，knuth的论文中有更加复杂的图案。
建图：行代表以一个格子为起点，使用第i个小矩形，横着或者竖着填充大矩形。
列代表每个格子。

sgu_433
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// 返回一组解，不是最优解
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#include <stdio.h>
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#include <memory.h>
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#include <iostream>
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#include <cstring>
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using namespace std;
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int dx[ ] =

{ -1, 0, 0, 1 };
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int dy[ ] =

{ 0, -1, 1, 0 };
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const int maxn = 1010;
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const int maxm = 105;
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const int maxt = maxn * maxm;
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int S[ maxm ], O[ maxn ]; // S[] 列链表中结点的总数 O[] 记录搜索结果
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int L[ maxt ], R[ maxt ], U[ maxt ], D[ maxt ]; // 四个方向
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int C[ maxt ], W[ maxt ]; // C[]列指针头 W[]行指针头
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int mat[maxn][maxm]; // 稀疏矩阵
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int ANS, CNT;
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void build( int n, int m ) // 邻接矩阵的build函数
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{
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int i, j, tot = m, first;
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R[0] = 1; L[0] = m;
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for( i = 1; i <= m; i++ )
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{
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L[i] = i - 1; R[i] = ( i + 1 ) % ( m + 1 );
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U[i] = D[i] = C[i] = i; S[i] = 0;
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}
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for( i = 1; i <= n; i++ )
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{
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first = 0;
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for( j = 1; j <= m; j++ )if( mat[i][j] )
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{
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tot++;
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U[tot] = U[j]; D[tot] = j;
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D[U[j]] = tot; U[j] = tot;
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if( first == 0 ) first = L[tot] = R[tot] = tot;
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else
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{
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L[tot] = L[first]; R[tot] = first;
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R[L[first]] = tot; L[first] = tot;
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}
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W[tot] = i; C[tot] = j; S[j]++;
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}
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}
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}
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void remove( int c )
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{
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int i, j;
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//remove column c and all row i that A[i][c] == 1
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L[R[c]] = L[c];
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R[L[c]] = R[c];
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//remove column c;
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for( i = D[c]; i != c; i = D[i] )
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{
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//remove i that A[i][c] == 1
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for( j = R[i]; j != i; j = R[j] )
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{
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U[D[j]] = U[j];
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D[U[j]] = D[j];
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S[C[j]]--;
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//decrease the count of column C[j];
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}
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}
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}
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void resume( int c )
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{
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int i, j;
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for( i = U[c]; i != c; i = U[i] )
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{
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for( j = L[i]; j != i; j = L[j] )
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{
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S[C[j]]++;
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U[D[j]] = j;
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D[U[j]] = j;
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}
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}
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L[R[c]] = c;
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R[L[c]] = c;
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}
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int dfs( int k )
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{
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CNT++;
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if( CNT > 10000 ) return 0;
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int i, j, t, m = maxm, mn;
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if( R[0] == 0 )
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{
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//One of the answers has been found.
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ANS = k;
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return 1;
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}
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for( t = R[0]; t != 0; t = R[t] )
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if( S[t] < m )

{ m = S[t]; mn = t; }
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remove( mn );
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for( i = D[mn]; i != mn; i = D[i] )
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{
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O[k] = i;//record the answer.
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for( j = R[i]; j != i; j = R[j] )
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remove( C[j] );
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if( dfs( k + 1 ) ) return 1;
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for( j = L[i]; j != i; j = L[j] )
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resume( C[j] );
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}
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resume( mn );
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return 0;
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}
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int l[ 10 ];
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int ch[ 25 ][ 25 ];
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int vis[ 10 ];
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char col[ 30 ];
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int main( )
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{
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//freopen("out.out","w",stdout);
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int n, m, k, i, j, c;
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for( int i = 1; i <= 26; i++ )
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col[ i ] = i + 'a' - 1;
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while( scanf("%d%d%d",&n,&m,&k) != EOF )
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{
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for( int i = 0; i < k; i++ )
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scanf("%d",&l[ i ]);
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memset( mat, 0, sizeof( mat ) );
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int q = 1;
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for( int i = 0; i < n; i++ )
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{
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for( int j = 0; j < m; j++ )
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{
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for( int p = 0; p < k; p++ )
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{
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if( j + l[ p ] <= m ) // 横着
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for( int x = 0; x < l[ p ]; x++ )
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mat[ q ][ i * m + j + x + 1 ] = 1;
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q++;
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if( i + l[ p ] <= n ) // 竖着
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for( int x = 0; x < l[ p ]; x++ )
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mat[ q ][ ( i + x ) * m + j + 1 ] = 1;
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q++;
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}
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}
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}
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build( q - 1, n * m );
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CNT = 0;
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int z, x, y, u, v, xx, yy;
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if( dfs( 0 ) == 0 ) puts("NO");
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else
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{
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printf("YES\n");
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memset( ch, 0, sizeof( ch ) );
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for( int i = 0; i < ANS; i++ )
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{
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z = W[ O[ i ] ] - 1;
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x = z / ( 2 * k ) / m;
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y = z / ( 2 * k ) % m;
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u = z % ( 2 * k ) / 2; // 矩形类型
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v = z % 2; // 横or竖
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memset( vis, 0, sizeof( vis ) );
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int j;
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if( !v ) // 横着
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{
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for( int p = 0; p < l[ u ]; p++ )
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{
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for( j = 0; j < 4; j++ )
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{
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xx = x + dx[ j ], yy = y + p + dy[ j ];
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if( xx < n && xx >= 0 && yy < m && yy >= 0 )
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vis[ ch[ xx ][ yy ] ] = 1;
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}
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}
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for( j = 1; ; j++ ) if( !vis[ j ] ) break;
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for( int p = 0; p < l[ u ]; p++ ) ch[ x ][ p + y ] = j;
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}
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else // 竖着
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{
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for( int p = 0; p < l[ u ]; p++ )
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{
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for( j = 0; j < 4; j++ )
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{
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xx = x + p + dx[ j ], yy = y + dy[ j ];
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if( xx < n && xx >= 0 && yy < m && yy >= 0 )
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vis[ ch[ xx ][ yy ] ] = 1;
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}
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}
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for( j = 1; ; j++ ) if( !vis[ j ] ) break;
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for( int p = 0; p < l[ u ]; p++ ) ch[ p + x ][ y ] = j;
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}
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}
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for( int i = 0; i < n; i++ )
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{
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for( int j = 0; j < m; j++ )
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printf("%c",col[ ch[ i ][ j ] ]);
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printf("\n");
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}
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}
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}
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return 0;
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}
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sgu 433 Japhshan and Ramshut

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