【COGS】1535 [ZJOI2004]树的果实 树状数组

传送门：【COGS】1535 [ZJOI2004]树的果实

题目分析：遇到和树有关的肯定先dfs，首先将美味值离散化（反正所有的都不相同，随便离散），然后按照题目给的树dfs，顺便记录其时间戳用以求前两个答案，因为dfs走的恰好是从根到一个点的路径，所以在dfs是顺带用树状数组维护果实的个数，答案三就是查询前面添加的果实有多少比当前的大就行了。

接下来把果实按照美味值从大到小排序（其实还是同一个比较函数，到过来读取就行了），每次为这个果实 u 查找子树区间[ in[ u ] , ou[ u ] ]内有多少比他大就好了，这个是答案一，然后答案二就等于n - fruit[ u ].x(x为离散后的)-答案一。

是不是很简单？就是我写的比较搓，跑的很慢。

代码如下：

#include <cmath>#include <cstdio>#include <cstring>#include <algorithm>using namespace std ;#define REP( i , a , b ) for ( int i = ( a ) ; i <  ( b ) ; ++ i )#define FOR( i , a , b ) for ( int i = ( a ) ; i <= ( b ) ; ++ i )#define REV( i , a , b ) for ( int i = ( a ) ; i >= ( b ) ; -- i )#define travel( e , H , u ) for ( Edge* e = H[u] ; e ; e = e -> next )#define CLR( a , x ) memset ( a , x , sizeof a )#define CPY( a , x ) memcpy ( a , x , sizeof a )#define ls ( o << 1 )#define rs ( o << 1 | 1 )#define lson ls , l , m#define rson rs , m + 1 , r#define mid ( ( l + r ) >> 1 )#define root 1 , 1 , n#define rt o , l , rconst int MAXN = 100005 ;const int MAXE = 100005 ;struct Edge {int v ;Edge* next ;} E[MAXE] , *H[MAXN] , *cur ;struct Node {int x , idx ;} fruit[MAXN] ;int in[MAXN] , ou[MAXN] , dfs_clock ;int ans1[MAXN] , ans2[MAXN] , ans3[MAXN] ;int T[MAXN] ;int n ;int cmp1 ( const Node& a , const Node& b ) {return a.x < b.x ;}int cmp2 ( const Node& a , const Node& b ) {return a.idx < b.idx ;}void add ( int x , int v ) {while ( x ) {T[x] += v ;x -= x & -x ;}}int sum ( int x , int ans = 0 ) {if ( !x ) return 0 ;while ( x <= n ) {ans += T[x] ;x += x & -x ;}return ans ;}void clear () {cur = E ;dfs_clock = 0 ;CLR ( H , 0 ) ;CLR ( T , 0 ) ;}void addedge ( int u , int v ) {cur -> v = v ;cur -> next = H[u] ;H[u] = cur ++ ;}void dfs ( int u ) {in[u] = ++ dfs_clock ;ans3[u] = sum ( fruit[u].x ) ;add ( fruit[u].x ,  1 ) ;travel ( e , H , u ) dfs ( e -> v ) ;add ( fruit[u].x , -1 ) ;ou[u] = dfs_clock ;}void solve () {int x ;clear () ;scanf ( "%d" , &n ) ;FOR ( i , 2 , n ) {scanf ( "%d" , &x ) ;addedge ( x , i ) ;}FOR ( i , 1 , n ) {scanf ( "%d" , &fruit[i].x ) ;fruit[i].idx = i ;}sort ( fruit + 1 , fruit + n + 1 , cmp1 ) ;FOR ( i , 1 , n ) fruit[i].x = i ;sort ( fruit + 1 , fruit + n + 1 , cmp2 ) ;dfs ( 1 ) ;sort ( fruit + 1 , fruit + n + 1 , cmp1 ) ;CLR ( T , 0 ) ;REV ( i , n , 1 ) {int idx = fruit[i].idx ;//printf ( "%d %d\n" , ou[idx] , in[idx] - 1 ) ;ans1[idx] = sum ( in[idx] ) - sum ( ou[idx] + 1 ) ;ans2[idx] = n - fruit[i].x - ans1[idx] ;add ( in[idx] , 1 ) ;}FOR ( i , 1 , n ) printf ( "%d %d %d\n" , ans1[i] , ans2[i] , ans3[i] ) ;}int main () {freopen ( "treesfruits.in" , "r" , stdin ) ;freopen ( "treesfruits.out" , "w" , stdout ) ;solve () ;}