标签:sse 没有 using 题意 位置 rhs cst span inline
题目传送门:ヾ(?ω?`)o
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。
#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;
}
【题解】Berland.Taxi Codeforces 883L 模拟 线段树 堆
标签:sse 没有 using 题意 位置 rhs cst span inline
原文地址:http://www.cnblogs.com/mlystdcall/p/7900621.html