【题解】Berland.Taxi Codeforces 883L 模拟 线段树 堆

Prelude

题目传送门：ヾ(•ω•`)o

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Solution

按照题意模拟即可。
维护一个优先队列，里面装的是正在运营中的出租车，关键字是乘客的下车时间。
维护一个线段树，第\(i\)个位置表示第\(i\)个房子前面有没有停放出租车，这样在有人需要打车的时候可以快速找到离她最近的车的位置。
对每个房子维护一个堆，里面装的是停在这个房子前面的出租车，关键字是出租车的编号和上一个乘客下车的时间，上一个乘客下车越早，等待时间越长。
然后模拟时间的流逝就可以了，代码非常好写。

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Code

#include <cstring>
#include <algorithm>
#include <cstdio>
#include <queue>
#include <utility>
#include <cstdlib>
#include <cassert>

using namespace std;
typedef long long ll;
const int MAXN = 200010;
const int INF = 0x3f3f3f3f;
int _w;

int n, k, m, x[MAXN], a[MAXN], b[MAXN];
ll t[MAXN];

namespace SGT {
	int sumv[MAXN<<2], ql, qr, qv;
	void _add( int o, int L, int R ) {
		sumv[o] += qv;
		if( L == R ) return;
		int M = (L+R)/2, lc = o<<1, rc = lc|1;
		if( ql <= M ) _add(lc, L, M);
		else _add(rc, M+1, R);
	}
	void add( int p, int v ) {
		ql = p, qv = v;
		_add(1, 1, n);
	}
	void _queryl( int o, int L, int R ) {
		if( L >= ql && R <= qr ) {
			if( !sumv[o] ) return;
			while( L != R ) {
				int M = (L+R)/2, lc = o<<1, rc = lc|1;
				if( sumv[lc] ) o = lc, R = M;
				else o = rc, L = M+1;
			}
			qv = L;
		} else {
			int M = (L+R)/2, lc = o<<1, rc = lc|1;
			if( ql <= M && !qv ) _queryl(lc, L, M);
			if( qr > M && !qv ) _queryl(rc, M+1, R);
		}
	}
	int queryl( int l, int r ) {
		ql = l, qr = r, qv = 0;
		_queryl(1, 1, n);
		return qv;
	}
	void _queryr( int o, int L, int R ) {
		if( L >= ql && R <= qr ) {
			if( !sumv[o] ) return;
			while( L < R ) {
				int M = (L+R)/2, lc = o<<1, rc = lc|1;
				if( sumv[rc] ) o = rc, L = M+1;
				else o = lc, R = M;
			}
			qv = L;
		} else {
			int M = (L+R)/2, lc = o<<1, rc = lc|1;
			if( qr > M && !qv ) _queryr(rc, M+1, R);
			if( ql <= M && !qv ) _queryr(lc, L, M);
		}
	}
	int queryr( int l, int r ) {
		ql = l, qr = r, qv = 0;
		_queryr(1, 1, n);
		return qv;
	}
}

struct Node {
	ll t;
	int id;
	Node() {}
	Node( ll t, int id ):
		t(t), id(id) {}
	bool operator<( const Node &rhs ) const {
		return t == rhs.t ? id > rhs.id : t > rhs.t;
	}
};
priority_queue<Node> pq[MAXN], evt;

void prelude() {
	for( int i = 1; i <= k; ++i ) {
		pq[x[i]].push( Node(0, i) );
		SGT::add(x[i], 1);
	}
}

void run( ll t ) {
	while( !evt.empty() && evt.top().t <= t ) {
		Node car = evt.top(); evt.pop();
		SGT::add(x[car.id], 1);
		// printf( "car.id = %d, x[id] = %d\n", car.id, x[car.id] );
		pq[x[car.id]].push(car);
	}
}
int use( int pos ) {
	Node car = pq[pos].top(); pq[pos].pop();
	SGT::add(pos, -1);
	return car.id;
}
int freecar( int pos ) {
	if( !SGT::sumv[1] ) return 0;
	int left = SGT::queryr(1, pos);
	int right = SGT::queryl(pos, n);
	// printf( "left = %d, right = %d\n", left, right );
	if( !left ) left = -INF;
	if( !right ) right = INF;
	if( left == right ) {
		return use(left);
	} else if( pos-left < right-pos ) {
		return use(left);
	} else if( right-pos < pos-left ) {
		return use(right);
	} else {
		Node cl = pq[left].top(), cr = pq[right].top();
		if( cl.t == cr.t ) {
			if( cl.id < cr.id ) {
				return use(left);
			} else {
				return use(right);
			}
		} else if( cl.t < cr.t ) {
			return use(left);
		} else {
			return use(right);
		}
	}
}
void solve() {
	ll now = 0;
	for( int i = 0; i < m; ++i ) {
		now = max(now, t[i]);
		run(now);
		int car = freecar( a[i] );
		// printf( "now = %lld, car = %d\n", now, car );
		if( !car ) {
			now = max(now, evt.top().t);
			run(now);
			// printf( "now = %lld, car = %d\n", now, car );
			car = freecar( a[i] );
		}
		// printf( "now = %lld, car = %d\n", now, car );
		printf( "%d %lld\n", car, now - t[i] + abs(x[car] - a[i]) );
		evt.push( Node(now + abs(x[car] - a[i]) + abs(a[i] - b[i]), car) );
		x[car] = b[i];
	}
}

int main() {
	_w = scanf( "%d%d%d", &n, &k, &m );
	for( int i = 1; i <= k; ++i )
		_w = scanf( "%d", x+i );
	for( int i = 0; i < m; ++i )
		_w = scanf( "%lld%d%d", t+i, a+i, b+i );
	prelude(), solve();
	return 0;
}