hdu 2871 Memory Control(伸展树splay tree)
题意:就是对一个区间的四种操作,NEW x,占据最左边的连续的x个单元,Free x 把x单元所占的连续区间清空 , Get x 把第x次占据的区间输出来, R 清空整个区间。
解题思路:这个题就是一个区间合并,以前用线段树写的,拿来练练splay。要记录的是区间最大的连续空格,要维护这个最值,需要两个辅助的值,该区间左边连续的最值和右边连续的最值。更新的时候仔细就好了,其他就是splay的常规操作的。还有就是记录占据的连续区间和查找占据的连续区间用个vector,二分查找,插入就好了。在清空整个区间时,用update,不要重建一棵树。
(看似简单,我调了一天啊。。起初是push_up写不好,调很久没弄出来,后来过了样例,一直TLE。。还以为是效率不够高,第二天才发现是内存池的tot没清零)
#pragma comment(linker, "/STACK:1024000000,1024000000") #include<stdio.h> #include<string.h> #include<algorithm> #include<vector> using namespace std ; vector< pair<int,int> > vec ; const int maxn = 55555 ; int tot , n , m ; int lm[maxn] , rm[maxn] , mm[maxn] , col[maxn] , val[maxn] ; int son[2][maxn] , fa[maxn] , size[maxn] , pos[maxn] ; void new_node ( int l , int r ) { size[tot] = 1 ; son[0][tot] = son[1][tot] = fa[tot] = -1 ; lm[tot] = rm[tot] = mm[tot] = r - l + 1 ; col[tot] = 1 , val[tot] = 1 ; if ( l == 0 ) col[tot] = -1 , lm[tot] = 0 , rm[tot] -- , mm[tot] -- ; if ( r == n + 1 ) col[tot] = -1 , rm[tot] = 0 , lm[tot] -- , mm[tot] -- ; if ( ( ( l + r ) >> 1 == 0 ) || ( ( l + r ) >> 1 == n + 1 ) ) val[tot] = 0 ; tot ++ ; } void push_down ( int rt ) { if ( col[rt] != -1 ) { int ls = son[0][rt] , rs = son[1][rt] ; val[ls] = val[rs] = col[rt] ; if ( ls != -1 ) { col[ls] = col[rt] ; lm[ls] = rm[ls] = mm[ls] = size[ls] * col[rt] ; } if ( rs != -1 ) { col[rs] = col[rt] ; lm[rs] = rm[rs] = mm[rs] = size[rs] * col[rt] ; } col[rt] = -1 ; } } void push_up ( int rt ) { size[rt] =1 ; mm[rt] = lm[rt] = rm[rt] = val[rt] ; int ls = son[0][rt] , rs = son[1][rt] ; if ( val[rt] ) { if ( ls == -1 ) lm[rt] = 1 + ( rs == -1 ? 0 : lm[rs] ) ; if ( rs == -1 ) rm[rt] = 1 + ( ls == -1 ? 0 : rm[ls] ) ; } if ( ls != -1 ) { lm[rt] = lm[ls] ; mm[rt] = max ( mm[ls] , rm[ls] + val[rt] ) ; if ( lm[ls] == size[ls] ) lm[rt] += val[rt] , mm[rt] += val[rt] ; size[rt] += size[ls] ; } if ( rs != -1 ) { rm[rt] = rm[rs] ; mm[rt] = max ( mm[rt] , max ( mm[rs] , lm[rs] + val[rt] ) ) ; if ( rm[rs] == size[rs] ) rm[rt] += val[rt] , mm[rt] = max ( mm[rt] , rm[rs] + val[rt] ) ; size[rt] += size[rs] ; } if ( ls != -1 && rs != -1 && val[rt] ) { mm[rt] = max ( mm[rt] , rm[ls] + lm[rs] + 1 ) ; if ( lm[ls] == size[ls] ) lm[rt] = lm[ls] + 1 + lm[rs] ; if ( rm[rs] == size[rs] ) rm[rt] = rm[rs] + 1 + rm[ls] ; } } int build ( int l , int r ) { if ( l > r ) return -1 ; int mid = ( l + r ) >> 1 ; new_node ( l , r ) ; int temp = tot - 1 ; son[0][temp] = build ( l , mid - 1 ) ; if ( son[0][temp] != -1 ) fa[son[0][temp]] = temp , size[temp] += size[son[0][temp]] ; son[1][temp] = build ( mid + 1 , r ) ; if ( son[1][temp] != -1 ) fa[son[1][temp]] = temp , size[temp] += size[son[1][temp]] ; return temp ; } void rot ( int rt , int c ) { int y = fa[rt] , z = fa[y] ; push_down ( y ) , push_down ( rt ) ; son[!c][y] = son[c][rt] ; if ( son[c][rt] != -1 ) fa[son[c][rt]] = y ; fa[rt] = z ; if ( z != -1 ) { if ( y == son[0][z] ) son[0][z] = rt ; else son[1][z] = rt ; } son[c][rt] = y , fa[y] = rt ; push_up ( y ) ; } void splay ( int rt , int to ) { push_down ( rt ) ; while ( fa[rt] != to ) { if ( fa[fa[rt]] == to ) rot ( rt , rt == son[0][fa[rt]] ) ; else { int y = fa[rt] , z = fa[y] ; if ( rt == son[0][y] ) { if ( y == son[0][z] ) rot ( y , 1 ) , rot ( rt , 1 ) ; else rot ( rt , 1 ) , rot ( rt , 0 ) ; } else { if ( y == son[1][z] ) rot ( y , 0 ) , rot ( rt , 0 ) ; else rot ( rt , 0 ) , rot ( rt , 1 ) ; } } } push_up ( rt ) ; } int find ( int rt , int key ) { int cnt = 0 ; if ( son[0][rt] != -1 ) cnt = size[son[0][rt]] ; if ( cnt + 1 == key ) return rt ; if ( cnt >= key ) return find ( son[0][rt] , key ) ; return find ( son[1][rt] , key - cnt - 1 ) ; } int update ( int l , int r , int c , int rt ) { l ++ , r ++ ; int temp = pos[l-1] ; splay ( temp , -1 ) ; rt = temp ; temp = pos[r+1] ; splay ( temp , rt ) ; temp = son[0][temp] ; col[temp] = val[temp] = c ; lm[temp] = rm[temp] = mm[temp] = size[temp] * c ; push_up ( fa[temp] ) ; push_up ( fa[fa[temp]] ) ; return rt ; } int search ( int rt , int key , int rk ) { push_down ( rt ) ; int ls = son[0][rt] , rs = son[1][rt] ; if ( ls != -1 && mm[ls] >= key ) { return search ( ls , key , rk ) ; } if ( val[rt] ) { int cnt = 1 ; if ( ls != -1 ) cnt += rm[ls] ; if ( rs != -1 ) cnt += lm[rs] ; if ( cnt >= key ) { int pos = rk ; if ( ls != -1 ) pos += size[ls] ; return pos - rm[ls] ; } } return search ( rs , key , rk + 1 + ( ls == -1 ? 0 : size[ls] ) ) ; } int bin ( int key ) { int l = 0 , r = vec.size () - 1 ; while ( l <= r ) { int m = ( l + r ) >> 1 ; if ( vec[m].second >= key ) r = m - 1 ; else l = m + 1 ; } return r + 1 ; } int bin2 ( int key ) { int l = 0 , r = vec.size () - 1 ; while ( l <= r ) { int m = ( l + r ) >> 1 ; if ( vec[m].second <= key ) l = m + 1 ; else r = m - 1 ; } return l ; } int get_num() { char a ; int num = 0 ; int flag = 1 ; while ( a = getchar() , ( a < '0' || a > '9' ) && a != '-' ) ; if ( a == '-' ) flag = -1 ; else num = a - '0' ; while (( a = getchar()) != ' ' && a != '\n' ) num = num * 10 + (a-'0') ; return num * flag ; } int main () { char op[11] ; int a , b , i ; while ( scanf ( "%d%d" , &n , &m ) != EOF ) { tot = 0 ; int root = build ( 0 , n + 1 ) ; for ( i = 1 ; i <= n + 2 ; i ++ ) pos[i] = find ( root , i ) ; pair<int,int> u ; vec.clear () ; while ( m -- ) { scanf ( "%s" , op ) ; if ( op[0] == 'R' ) { root = update ( 1 , n , 1 , root ) ; vec.clear () ; puts ( "Reset Now" ) ; } else if ( op[0] == 'N' ) { a = get_num () ; if ( mm[root] < a ) puts ( "Reject New" ) ; else { int l = search ( root , a , 0 ) ; printf ( "New at %d\n" , l ) ; int r = l + a - 1 ; root = update ( l , r , 0 , root ) ; u = make_pair ( l , r ) ; l = bin2 ( r ) ; vec.insert ( vec.begin () + l , u ) ; } } else if ( op[0] == 'F' ) { a = get_num () ; int k = bin ( a ) ; if ( k == vec.size () || vec[k].first > a ) puts ( "Reject Free" ) ; else { printf ( "Free from %d to %d\n" , vec[k].first , vec[k].second ) ; root = update ( vec[k].first , vec[k].second , 1 , root ) ; vec.erase ( vec.begin () + k , vec.begin () + k + 1 ) ; } } else { a = get_num () ; if ( a > vec.size () ) puts ( "Reject Get" ) ; else printf ( "Get at %d\n" , vec[a-1].first ) ; } } puts ( "" ) ; } } /* 10 2 N 9 N 1 */