传送门:【HDU】3663 Power Stations
题目分析:可以将每个镇子按照天数拆成N*D列,然后对于每个镇子的可行区间,按照区间的可行方案依次拆成行,这样本题就转化成了所有列的1的覆盖的问题,但是我们要保证每个镇子只能选一次,那么我们再增加N列,所有属于一个镇子x的区间,在N*D+x的位置设为1。那么就是Dancing Links的活了~
PS:找了一晚上的RE,真心纠结啊TUT,回到寝室用手机看代码,一直没发现哪里错。。。。然后最后一不小心瞟到输入,发现没加scanf != EOF 。。。。然后手机上加上个'~'交一下就AC了TUT。。。。
代码如下:
#include <cstdio> #include <cstring> #include <algorithm> using namespace std ; #define REP( i , a , b ) for ( int i = a ; i < b ; ++ i ) #define REV( i , a , b ) for ( int i = a - 1 ; i >= b ; -- i ) #define FOR( i , a , b ) for ( int i = a ; i <= b ; ++ i ) #define FOV( i , a , b ) for ( int i = a ; i >= b ; -- i ) #define REC( i , A , o ) for ( int i = A[o] ; i != o ; i = A[i] ) #define CLR( a , x ) memset ( a , x , sizeof a ) const int MAXN = 65 ; const int MAXM = 2000 ; const int MAXNODE = 500000 ; struct Node { int l , r , idx ; Node () {} Node ( int l , int r , int idx ) : l ( l ) , r ( r ) , idx ( idx ) {} } ; struct DLX { int U[MAXNODE] , D[MAXNODE] , L[MAXNODE] , R[MAXNODE] ; int row[MAXNODE] , col[MAXNODE] ; int S[MAXM] , H[MAXM] ; int deep , ans[MAXN] ; int n , m ; int size ; int N , M , DD ; int X[MAXN] , Y[MAXN] ; Node node[MAXM] ; int G[MAXN][MAXN] ; void remove ( int c ) { L[R[c]] = L[c] ; R[L[c]] = R[c] ; REC ( i , D , c ) REC ( j , R , i ) { D[U[j]] = D[j] ; U[D[j]] = U[j] ; -- S[col[j]] ; } } void resume ( int c ) { REC ( i , U , c ) REC ( j , L , i ) { ++ S[col[j]] ; U[D[j]] = j ; D[U[j]] = j ; } R[L[c]] = c ; L[R[c]] = c ; } int dance ( int d ) { if ( R[0] == 0 ) { deep = d ; return 1 ; } int c = R[0] ; REC ( i , R , 0 ) if ( S[c] > S[i] ) c = i ; //printf ( "ok\n" ) ; remove ( c ) ; REC ( i , D , c ) { ans[d] = row[i] ; REC ( j , R , i ) remove ( col[j] ) ; if ( dance ( d + 1 ) ) return 1 ; REC ( j , L , i ) resume ( col[j] ) ; } resume ( c ) ; return 0 ; } void link ( int r , int c ) { ++ size ; ++ S[c] ; row[size] = r ; col[size] = c ; U[size] = U[c] ; D[size] = c ; D[U[c]] = size ; U[c] = size ; if ( ~H[r] ) { R[size] = H[r] ; L[size] = L[H[r]] ; R[L[size]] = size ; L[R[size]] = size ; } else H[r] = L[size] = R[size] = size ; } void init () { CLR ( H , -1 ) ; FOR ( i , 0 , n ) { S[i] = 0 ; L[i] = i - 1 ; R[i] = i + 1 ; U[i] = i ; D[i] = i ; } L[0] = n ; R[n] = 0 ; size = n ; } void solve () { int x , y ; n = N * DD + N ; m = 0 ; init () ; CLR ( G , 0 ) ; CLR ( node , 0 ) ; REP ( i , 0 , M ) { scanf ( "%d%d" , &x , &y ) ; G[x][y] = G[y][x] = 1 ; } FOR ( i , 1 , N ) G[i][i] = 1 ; FOR ( idx , 1 , N ) { scanf ( "%d%d" , &x , &y ) ; int tmp = N * DD + idx ; ++ m ; link ( m , tmp ) ;//none select node[m] = Node ( 0 , 0 , idx ) ; FOR ( i , x , y ) FOR ( j , i , y ) { ++ m ; link ( m , tmp ) ; node[m] = Node ( i , j , idx ) ; FOR ( a , 1 , N ) if ( G[idx][a] ) FOR ( b , i , j ) link ( m , ( a - 1 ) * DD + b ) ; } } if ( !dance ( 0 ) ) printf ( "No solution\n" ) ; else { CLR ( X , 0 ) ; CLR ( Y , 0 ) ; REP ( i , 0 , deep ) { X[node[ans[i]].idx] = node[ans[i]].l ; Y[node[ans[i]].idx] = node[ans[i]].r ; } FOR ( i , 1 , N ) printf ( "%d %d\n" , X[i] , Y[i] ) ; } printf ( "\n" ) ; } } dlx ; int main () { while ( ~scanf ( "%d%d%d" , &dlx.N , &dlx.M , &dlx.DD ) ) dlx.solve () ; return 0 ; }