传送门:【COGS】257 动态排名系统
题目分析:树状数组的每个节点就是一棵线段树,树状数组用以维护前缀和!主席树的区间加减求区间第K小。和zoj2112基本一样。
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 REV( i , a , b ) for ( int i = a ; i >= b ; -- i ) #define FOR( i , a , b ) for ( int i = a ; i <= b ; ++ i ) #define CLR( a , x ) memset ( a , x , sizeof a ) #define CPY( a , x ) memcpy ( a , x , sizeof a ) const int MAXN = 50005 ; const int MAXQ = 10005 ; const int MAXM = 60005 ; struct Tree { int ls , rs ; int s ; } T[10000000] ; struct Query { char op ; int l , r , k ; } Q[MAXQ] ; int newnode ; int a[MAXM] , cnt ; int root[MAXN] ; int use[MAXN] ; int num[MAXN] ; int n , q ; int unique ( int n ) { int cnt = 1 ; sort ( a + 1 , a + n + 1 ) ; FOR ( i , 2 , n ) if ( a[i] != a[cnt] ) a[++ cnt] = a[i] ; return cnt ; } int hash ( int x , int l = 1 , int r = cnt ) { while ( l < r ) { int m = ( l + r ) >> 1 ; if ( a[m] >= x ) r = m ; else l = m + 1 ; } return l ; } void build ( int& o , int l , int r ) { o = ++ newnode ; T[o].s = 0 ; if ( l == r ) return ; int m = ( l + r ) >> 1 ; build ( T[o].ls , l , m ) ; build ( T[o].rs , m + 1 , r ) ; } int insert ( int old , int pos , int val ) { int now = ++ newnode , tmp = now ; int l = 1 , r = cnt ; T[now].s = T[old].s + 1 ; while ( l < r ) { int m = ( l + r ) >> 1 ; if ( pos <= m ) { T[now].ls = ++ newnode ; T[now].rs = T[old].rs ; now = T[now].ls ; old = T[old].ls ; r = m ; } else { T[now].ls = T[old].ls ; T[now].rs = ++ newnode ; now = T[now].rs ; old = T[old].rs ; l = m + 1 ; } T[now].s = T[old].s + val ; } return tmp ; } void add ( int x , int pos , int v ) { while ( x <= n ) { root[x] = insert ( root[x] , pos , v ) ; x += x & -x ; } } int sum ( int x , int ans = 0 ) { while ( x ) { ans += T[T[use[x]].ls].s ; x -= x & -x ; } return ans ; } void build_tree () { build ( root[0] , 1 , cnt ) ; FOR ( i , 1 , n ) root[i] = root[0] ; FOR ( i , 1 , n ) add ( i , hash ( num[i] ) , 1 ) ; } int query ( int L , int R , int kth ) { for ( int i = L - 1 ; i ; i -= i & -i ) use[i] = root[i] ; for ( int i = R ; i ; i -= i & -i ) use[i] = root[i] ; int l = 1 , r = cnt ; while ( l < r ) { int m = ( l + r ) >> 1 ; int tmp = sum ( R ) - sum ( L - 1 ) ; if ( tmp >= kth ) { r = m ; for ( int i = L - 1 ; i ; i -= i & -i ) use[i] = T[use[i]].ls ; for ( int i = R ; i ; i -= i & -i ) use[i] = T[use[i]].ls ; } else { l = m + 1 ; kth -= tmp ; for ( int i = L - 1 ; i ; i -= i & -i ) use[i] = T[use[i]].rs ; for ( int i = R ; i ; i -= i & -i ) use[i] = T[use[i]].rs ; } } return l ; } void solve () { newnode = 0 ; cnt = 0 ; scanf ( "%d%d" , &n , &q ) ; FOR ( i , 1 , n ) { scanf ( "%d" , &num[i] ) ; a[++ cnt] = num[i] ; } REP ( i , 0 , q ) { scanf ( " %c" , &Q[i].op ) ; if ( Q[i].op == 'Q' ) scanf ( "%d%d%d" , &Q[i].l , &Q[i].r , &Q[i].k ) ; else { scanf ( "%d%d" , &Q[i].l , &Q[i].k ) ; a[++ cnt] = Q[i].k ; } } cnt = unique ( cnt ) ; build_tree () ; REP ( i , 0 , q ) { if ( Q[i].op == 'Q' ) printf ( "%d\n" , a[query ( Q[i].l , Q[i].r , Q[i].k )] ) ; else { add ( Q[i].l , hash ( num[Q[i].l] ) , -1 ) ; add ( Q[i].l , hash ( Q[i].k ) , 1 ) ; num[Q[i].l] = Q[i].k ; } } } int main () { freopen ( "dynrank.in" , "r" , stdin ) ; freopen ( "dynrank.out" , "w" , stdout ) ; int T ; scanf ( "%d" , &T ) ; while ( T -- ) solve () ; return 0 ; }