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

Prelude

题目传送门:ヾ(•ω•`)o


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。


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;
}
posted @ 2017-11-26 21:47 mlystdcall 阅读(...) 评论(...) 编辑 收藏