题意：
给你一段内存单元，你可以对它进行一下操作：
开辟一个内存块，如果开辟成功返回该内存块最左边的值；
删除包含X的内存段元，如果成功输出该内存块的左右值；
找到第X快内存块，如果成功输出该内存块的最左边的值；
清除内存；
这个题的思想与是一样的；这题还要做一个处理的是建立一个链表把之前的内存块首尾点给存储起来，这样就容易对它进行操作；
其余的地方与Hotel一样，这里就不累叙了；
View Code #include#include #include #include using namespace std; class Node { public: int l,r,l_value,r_value; int cover,max; }; class List { public: int start,end; }; vector list; Node tree[250000]; int Max( int a, int b ) { return a > b ? a:b; } void Maketree( int l , int r, int cnt ) { tree[cnt].l = l; tree[cnt].r = r; tree[cnt].l_value = tree[cnt].r_value = tree[cnt].max = r - l +1; tree[cnt].cover = 0; if( l==r ) { return ; } int mid = ( l + r )>>1; Maketree( l , mid, cnt*2 ); Maketree( mid + 1 , r ,cnt*2+1 ); } void fun( int cnt ) { if( tree[cnt].cover == 0 ) tree[cnt].l_value = tree[cnt].r_value = tree[cnt].max = tree[cnt].r - tree[cnt].l +1; else tree[cnt].l_value = tree[cnt].r_value = tree[cnt].max = 0; } void Update( int l , int r , int cnt ,int num ) { if( l > tree[cnt].r || tree[cnt].l > r) return ; if( tree[cnt].l >= l &&tree[cnt].r <= r ) { tree[cnt].cover = num; fun( cnt ); return ; } if( tree[cnt].cover != -1 ) { tree[cnt*2].cover = tree[cnt*2+1].cover =tree[cnt].cover; fun( cnt*2 ); fun( cnt*2+1 ); tree[cnt].cover = -1; } Update( l , r, cnt*2 , num ); Update( l ,r ,cnt*2+1 , num ); if( tree[cnt*2].cover==tree[cnt*2+1].cover ) tree[cnt].cover = tree[cnt*2].cover; else tree[cnt].cover = -1; tree[cnt].l_value = tree[cnt*2].l_value + ( tree[cnt*2].cover==0?tree[cnt*2+1].l_value:0 ); tree[cnt].r_value = tree[cnt*2+1].r_value + ( tree[cnt*2+1].cover == 0? tree[cnt*2].r_value:0 ); tree[cnt].max = Max( Max( tree[cnt*2].max , tree[cnt*2+1].max ), tree[cnt*2].r_value + tree[cnt*2+1].l_value ); } int Query( int num, int cnt ) { if( tree[cnt].l_value >= num &&tree[cnt].cover ==0 ) { Update( tree[cnt].l, tree[cnt].l+num -1 , 1, 1 );// printf( "asfasfa" ); return tree[cnt].l; } if( tree[cnt].cover == 1) return 0; if( tree[cnt].max>=num && tree[cnt].cover == -1 ) { if( tree[cnt*2].max >= num ) { return Query( num , cnt*2 ); } else if( tree[cnt*2].r_value + tree[cnt*2+1].l_value >= num ) { int t = tree[cnt*2].r - tree[cnt*2].r_value; Update( t +1, t + num , 1 , 1 ); return t+1 ; } else return Query( num , cnt*2+1 ); } return 0; } int find( int n ) { int begin = 0 ; int end = list.size()-1; while( begin <= end ) { int mid = ( begin + end )>>1; if( list[mid].start > n ) end = mid -1; else begin = mid + 1; } return begin; } int main( ) { int n , m , num,left ,right; char str[10]; while( scanf( "%d%d",&n,&m )==2 ) { Maketree( 1 , n , 1 ); list.clear(); while( m-- ) { scanf( "%s" ,str ); if( str[0]=='R' ) { printf( "Reset Now\n" ); Update(1 , n , 1,0 ); list.clear(); } else { scanf( "%d",&num ); switch( str[0] ) { case 'N': left = Query( num ,1 ); if( left ==0 ) printf( "Reject New\n" ); else { right = find( left ); List t; t.start = left ; t.end = left + num -1; list.insert( list.begin() + right , t ); printf( "New at %d\n",left ); } break; case 'F': right = find( num ) -1; if( right ==-1||list[right].end < num ) printf( "Reject Free\n" ); else { printf( "Free from %d to %d\n",list[right].start , list[right].end ); Update( list[right].start, list[right].end ,1, 0); list.erase( list.begin() + right ); } break; case 'G': if( num > list.size() ) printf( "Reject Get\n" ); else printf( "Get at %d\n",list[num-1].start ); break; } } } } return 0; }