【HDU】5021 Revenge of kNN II 树状数组

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传送门：【HDU】5021 Revenge of kNN II

题目分析：【HDU】4995 Revenge of kNN的升级版，这次取消的K<=10的限制。

但是依旧可以做！

首先我们将点按照横坐标从小到大排序，然后对于每次查询，我们先二分距离mid，然后再二分查找在X-mid，X+mid里面有多少数，如果小于K则抬升下界，如果大于K+1则降低上界，如果等于K则直接更新，还有就是正好等于K+1的时候，看最两端到底哪个应该被排除。更新值以及区间求和用树状数组维护就好了。

题解所提及的只用一次二分的方法不会，所以渣渣我用了两次二分。

代码如下：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std ;typedef long long LL ;#define travel( e , H , u ) for ( Edge* e = H[u] ; e ; e = e -> next )#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 = 100005 ;struct Node {int x , idx ;double v ;bool operator < ( const Node& a ) const {return x < a.x ;}} p[MAXN] ;int idx[MAXN] ;int n , m ;double T[MAXN] ;void add ( int x , double v ) {while ( x <= n ) {T[x] += v ;x += x & -x ;}}double sum ( int x , double ans = 0 ) {while ( x ) {ans += T[x] ;x -= x & -x ;}return ans ;}int L_find ( int x ) {int l = 1 , r = n ;while ( l < r ) {int m = ( l + r ) >> 1 ;if ( p[m].x >= x ) r = m ;else l = m + 1 ;}return l ;}int R_find ( int x ) {int l = 1 , r = n ;while ( l < r ) {int m = ( l + r + 1 ) >> 1 ;if ( p[m].x <= x ) l = m ;else r = m - 1 ;}return r ;}void solve () {int x , k ;double ans = 0 ;clr ( T , 0 ) ;scanf ( "%d%d" , &n , &m ) ;FOR ( i , 1 , n ) {scanf ( "%d%lf" , &p[i].x , &p[i].v ) ;p[i].idx = i ;}sort ( p + 1 , p + n + 1 ) ;FOR ( i , 1 , n ) add ( i , p[i].v ) ;FOR ( i , 1 , n ) idx[p[i].idx] = i ;while ( m -- ) {scanf ( "%d%d" , &x , &k ) ;int l = 1 , r = p[n].x ;int v = idx[x] ;while ( l <= r ) {int mid = ( l + r ) >> 1 ;int L = L_find ( p[v].x - mid ) ;int R = R_find ( p[v].x + mid ) ;if ( R - L < k ) l = mid + 1 ;else if ( R - L > k + 1 ) r = mid - 1 ;else if ( R - L == k ) {double tmp = ( sum ( R ) - sum ( L - 1 ) - p[v].v ) / k ;ans += tmp ;add ( v , -p[v].v ) ;add ( v , tmp ) ;p[v].v = tmp ;break ;} else if ( R - L == k + 1 ) {if ( p[v].x - p[L].x == p[R].x - p[v].x ) {if ( p[L].idx < p[R].idx ) -- R ;else ++ L ;}else if ( p[v].x - p[L].x < p[R].x - p[v].x ) -- R ;else ++ L ;double tmp = ( sum ( R ) - sum ( L - 1 ) - p[v].v ) / k ;ans += tmp ;add ( v , -p[v].v ) ;add ( v , tmp ) ;p[v].v = tmp ;break ;}}}printf ( "%.3f\n" , ans ) ;}int main () {int T ;scanf ( "%d" , &T ) ;while ( T -- ) solve () ;return 0 ;}

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