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