Codeforces1045G

Codeforces1045G

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做法：按半径r从大到小枚举，对于每个q，枚举对应位置可能的q值，对每个q，维护出现的坐标x，每次查询半径内的已经出现的坐标的数目即可。需要实现一个插入单点加，查询区间和的操作，动态开点线段树即可。看来还是要学习一下pb_ds了。

#include <bits/stdc++.h>
typedef long long ll;
const int N = 1e5 + 7;
const ll lim = 1e9 + 7;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while( ! isdigit(c) ) { if(c == '-') f = -1; c = getchar(); }
    while( isdigit(c) ) { x = x*10 + c - '0'; c = getchar(); }
    return x * f;
}
inline void write( ll x ) {
    if( x >= 10 ) write( x / 10 );
    putchar( x % 10 + '0' );
}
using namespace std;
int n , k;
ll ans = 0;
struct node { ll x , r , q; } a[N];
bool cmp( node a , node b ) { return a.r > b.r; }
set< ll > hv;
struct Tree {
    int lson , rson; ll sum;
}T[ N * 100 ];
map< ll , int > rt;
int cnt = 0;
void ins( int &rt , ll l , ll r , ll p ) {
    if( ! rt ) rt = ++ cnt;
    ++ T[ rt ].sum;
    if( l == r ) return;
    ll mid = ( l + r ) >> 1;
    if( p <= mid ) ins( T[ rt ].lson , l , mid , p);
    else ins( T[ rt ].rson , mid + 1 , r , p );
}
ll ask( int rt , ll l , ll r , ll p ) {
    if( ! rt ) return 0;
    if( r <= p ) return T[ rt ].sum;
    ll mid = ( l + r ) >> 1;
    if( p <= mid ) return ask( T[rt].lson , l , mid , p );
    else if( p > mid ) return T[ T[ rt ].lson ].sum + ask( T[rt].rson , mid + 1 , r , p );
}
int main() {
    n = read(), k = read();
    for ( int i = 1 ; i <= n ; ++ i ) {
        a[ i ].x = read() , a[ i ].r = read() , a[ i ].q = read();
        hv.insert( a[ i ].q );
    }
    sort(a + 1 , a + n + 1 , cmp );
    int p = 0;
    for ( int i = 1 ; i <= n ; ++ i ) {
        for ( ll t = a[ i ].q - k; t <= a[ i ].q + k ; ++ t)
            if( hv.find(t) != hv.end() ) {
                ans +=  ask( rt[t] , 0 , lim , a[ i ].x + a[ i ].r );
                ans -= ask( rt[t] , 0 , lim , a[ i ].x - a[ i ].r - 1);
            }
        ins( rt[a[ i ].q] , 0 , lim , a[i].x );
    }
    write( ans ); putchar( '\n' );
    return 0;
}