【8.30校内测试】【找规律模拟】【DP】【二分+贪心】
2024-09-27 05:58:44
对于和规律或者数学有关的题真的束手无策啊QAQ
首先发现两个性质:
1、不管中间怎么碰撞,所有蚂蚁的相对位置不会改变,即后面的蚂蚁不会超过前面的蚂蚁或者落后更后面的蚂蚁。
2、因为所有蚂蚁速度一样,不管标号的话两只蚂蚁的碰撞相当于直接互相穿过,所以最初有多少蚂蚁方向向左,最后就有多少蚂蚁从左落下,向右同理。
总结一下又可以发现,比如有$cntl$只蚂蚁最初向左,$cntr$只蚂蚁最初向右,那么最后就是原位置的左边连续$cntl$只从左落下,原位置右边连续$cntr$只从右落下。我们将所有方向向左和向右的蚂蚁落下的时间分别排序,和原序列上一一对应即可。
#include<iostream>
#include<cstdio>
#include<algorithm>
#define LL long long
using namespace std; int n, b[], cntl, cntr;
LL a[], L;
double ans[], l[], r[]; int main ( ) {
freopen ( "ant.in", "r", stdin );
freopen ( "ant.out", "w", stdout );
scanf ( "%I64d%d", &L, &n );
for ( int i = ; i <= n; i ++ ) scanf ( "%I64d", &a[i] );
for ( int i = ; i <= n; i ++ ) scanf ( "%d", &b[i] );
for ( int i = ; i <= n; i ++ )
if ( !b[i] )
l[++cntl] = a[i];
else r[++cntr] = L - a[i];
sort ( l + , l + + cntl );
sort ( r + , r + + cntr );
for ( int i = ; i <= cntl; i ++ )
ans[i] = l[i];
for ( int i = ; i <= cntr; i ++ )
ans[n - i + ] = r[i];
for ( int i = ; i <= n; i ++ )
printf ( "%.2lf ", ans[i] );
return ;
}
见8.20校内测试,题目转换一下就一模一样了。
数据比较水,写的$O(nlog_nlog_h)$完全够了。
二分最小值,$check$的时候贪心修改区间,我用的线段树,判断一下就好了。实际上差分复杂度更优。写线段树的时候无聊写了区间求和??
#include<iostream>
#include<cstdio>
#include<algorithm>
#define LL long long
using namespace std; int n, K;
LL T, a[]; LL TR[], tag[], vc[]; void update ( int nd ) {
TR[nd] = TR[nd << ] + TR[nd << | ];
} void push_down ( int nd, int l, int r ) {
if ( tag[nd] ) {
int mid = ( l + r ) >> ;
TR[nd << ] += tag[nd] * ( mid - l + );
TR[nd << | ] += tag[nd] * ( r - mid );
tag[nd << ] += tag[nd];
tag[nd << | ] += tag[nd];
tag[nd] = ;
}
} void build ( int nd, int l, int r ) {
TR[nd] = ; tag[nd] = ;
if ( l == r ) {
TR[nd] = a[vc[l]];
return ;
}
int mid = ( l + r ) >> ;
build ( nd << , l, mid );
build ( nd << | , mid + , r );
update ( nd );
} void add ( int nd, int l, int r, int L, int R, LL d ) {
if ( l >= L && r <= R ) {
TR[nd] += ( r - l + ) * d;
tag[nd] += d;
return ;
}
push_down ( nd, l, r );
int mid = ( l + r ) >> ;
if ( L <= mid ) add ( nd << , l, mid, L, R, d );
if ( R > mid ) add ( nd << | , mid + , r, L, R, d );
update ( nd );
} LL query ( int nd, int l, int r, int pos ) {
if ( l == r ) return TR[nd];
push_down ( nd, l, r );
int mid = ( l + r ) >> ;
if ( pos <= mid ) return query ( nd << , l, mid, pos );
else return query ( nd << | , mid + , r, pos );
} bool check ( LL mid ) {
int tot = ; LL sum = ;
for ( int i = ; i <= n; i ++ )
if ( a[i] < mid ) vc[++tot] = i;
build ( , , tot );
vc[++tot] = 0x7f7f7f7f7f7f7f;
for ( int i = ; i < tot; i ++ ) {
LL now = query ( , , tot - , i );
if ( now >= mid ) continue;
if ( mid - now + sum > T ) { sum = T + ; break; }
int to = vc[i] + K - ;
int pos = upper_bound ( vc + , vc + + tot, to ) - vc - ;
add ( , , tot - , i, pos, mid - now );
sum += mid - now;
}
if ( sum <= T ) return ;
return ;
} LL MI = 0x3f3f3f3f, MA;
LL erfen ( ) {
LL l = MI, r = MA + T, ans;
while ( l <= r ) {
int mid = ( l + r ) >> ;
if ( check ( mid ) ) l = mid + , ans = mid;
else r = mid - ;
}
return ans;
} int main ( ) {
freopen ( "watering.in", "r", stdin );
freopen ( "watering.out", "w", stdout );
scanf ( "%d%d%I64d", &n, &K, &T );
for ( int i = ; i <= n; i ++ ) scanf ( "%I64d", &a[i] ), MI = min ( MI, a[i] ), MA = max ( MA, a[i] );
LL ans = erfen ( );
printf ( "%I64d", ans );
return ;
}
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