bzoj3133

3133: [Baltic2013]ballmachine

Time Limit: 20 Sec  Memory Limit: 128 MB
Submit: 108  Solved: 41
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Description

有一个装球机器，构造可以看作是一棵树。有下面两种操作：

• 从根放入一个球，只要下方有空位，球会沿着树滚下。如果同时有多个点可以走，那么会选择编号最小的节点所在路径的方向。比如依次在树根4放2个球，第一个球会落到1，第二个会落到3：
• 

• 从某个位置拿走一个球，那么它上方的球会落下来。比如依次拿走5, 7, 8三个球：

Input

第一行：球的个数N，操作个数Q （N, Q <= 100 000）下面N行：第i个节点的父亲。如果是根，则为0 接下来Q行：op num

1. op == 1：在根放入num个球
2. op == 2：拿走在位置num的球

Output

保证输入合法

1. op == 1：输出最后一个球落到了哪里
2. op == 2：输出拿走那个球后有多少个球会掉下来

Sample Input

8 4
0
1
2
2
3
3
4
6
1 8
2 5
2 7
2 8

Sample Output

1
3
2
2

HINT

Source

abcdabcd987提供

题解：

考虑从空树一直放球直到放满为止，则掉落位置构成一个全序。

比如样例中的顺序是58637421，那么当位置6没有球时，球就不会掉到37421上。

所以用堆维护所有球可能掉到的位置就可以了。这些位置就是没有球的点构成的树的叶子节点。

取出一个球就是找从这个点到根最浅的有球的点，可以用树剖或倍增。（树剖又慢又长。）

（想知道怎么写能更短。。。）

#include <stdio.h>#include <algorithm>#include <queue>using namespace std;const int maxn = 120000;int order[maxn];struct so {int v , minn;};struct getnext {int v;};bool operator > ( so x1 , so x2 ) {return x1.minn > x2.minn;}bool operator > ( getnext x1 , getnext x2 ) {return order[x1.v] > order[x2.v];}struct heap {priority_queue < getnext , vector < getnext > , greater < getnext > > p , q;getnext top () {while ( q.size () != 0 && p.top ().v == q.top ().v ) p.pop (), q.pop ();return p.top ();}bool empty () {while ( q.size () != 0 && p.top ().v == q.top ().v ) p.pop (), q.pop ();if ( p.size () == 0 ) return true;return false;}void push ( int x ) {getnext tmp;tmp.v = x;p.push ( tmp );}void pop ( int x ) {getnext tmp;tmp.v = x;q.push ( tmp );}} h;struct tree {int v;tree *next;} poolt[maxn*2] , *g[maxn];int topt;int n , q;int fa[maxn] , dep[maxn] , sonnum[maxn] , mnum[maxn];int f[maxn];int go[18][maxn];int root;int index1;void add ( int u , int v ) {tree *tmp = &poolt[++topt];tmp -> v = v; tmp -> next = g[u]; g[u] = tmp;}void dfs1 ( int i ) {mnum[i] = i;for ( tree *j = g[i] ; j ; j = j -> next ) if ( !mnum[j->v] ) {sonnum[i]++;dep[j->v] = dep[i] + 1;fa[j->v] = i;dfs1 ( j -> v );mnum[i] = min ( mnum[i] , mnum[j->v] );}}void getorder ( int i , int from ) {priority_queue < so , vector < so > , greater < so > > que;so tmp;for ( tree *j = g[i] ; j ; j = j -> next ) if ( j -> v != from ) {tmp.v = j -> v; tmp.minn = mnum[j->v];que.push ( tmp );}while ( que.size () != 0 ) {getorder ( que.top().v , i );que.pop ();}order[i] = ++index1;}void change ( int x ) {f[x] = 1 - f[x];}int find ( int i ) {for ( int j = 16 ; j >= 0 ; j-- ) if ( f[go[j][i]] == 1 ) i = go[j][i];return i;}void put () {getnext tmp = h.top ();h.pop ( tmp.v );change ( tmp.v );//printf ( "%d\n" , w[tmp.v] );sonnum[fa[tmp.v]]--;if ( sonnum[fa[tmp.v]] == 0 ) h.push ( fa[tmp.v] ); }int take ( int i ) {int tmp = find ( i );//printf ( "%d\n" , tmp );change ( tmp );h.push ( tmp ); if ( sonnum[fa[tmp]] == 0 ) h.pop ( fa[tmp] );sonnum[fa[tmp]]++; return dep[i] - dep[tmp];}void work () {int i , j , v , op , num , ans;scanf ( "%d%d" , &n , &q );for ( i = 1 ; i <= n ; i++ ) {scanf ( "%d" , &v );if ( v != 0 ) add ( i , v ), add ( v , i );else root = i;}dep[root] = 1;dfs1 ( root );for ( i = 1 ; i <= n ; i++ ) go[0][i] = fa[i];for ( i = 1 ; i <= 16 ; i++ ) for ( j = 1 ; j <= n ; j++ ) go[i][j] = go[i-1][go[i-1][j]];//for ( i = 1 ; i <= n ; i++ ) printf ( "%d\n" , w[i] );getorder ( root , -1 );for ( i = 1 ; i <= n ; i++ ) if ( sonnum[i] == 0 ) h.push ( i ); for ( i = 1 ; i <= q ; i++ ) {scanf ( "%d%d" , &op , &num );if ( op == 1 ) {for ( j = 1 ; j <= num ; j++ ) {ans = h.top ().v;//printf ( "%d %d %d\n" , h.top().v , h.top().minn , h.top().dep );put ();}printf ( "%d\n" , ans );}else {ans = take ( num );printf ( "%d\n" , ans );}}}int main () {//FILE *fpr = freopen ( "bzoj3133.in" , "r" , stdin );//FILE *fpw = freopen ( "bzoj3133.out" , "w" , stdout );work ();return 0;}