【8.30校内测试】【找规律模拟】【DP】【二分+贪心】

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[100005], cntl, cntr;
LL a[100005], L;
double ans[100005], l[100005], r[100005];

int main ( ) {
    freopen ( "ant.in", "r", stdin );
    freopen ( "ant.out", "w", stdout );
    scanf ( "%I64d%d", &L, &n );
    for ( int i = 1; i <= n; i ++ )    scanf ( "%I64d", &a[i] );
    for ( int i = 1; i <= n; i ++ )    scanf ( "%d", &b[i] );
    for ( int i = 1; i <= n; i ++ )
        if ( !b[i] )
            l[++cntl] = a[i];
        else r[++cntr] = L - a[i];
    sort ( l + 1, l + 1 + cntl );
    sort ( r + 1, r + 1 + cntr );
    for ( int i = 1; i <= cntl; i ++ )
        ans[i] = l[i];
    for ( int i = 1; i <= cntr; i ++ )
        ans[n - i + 1] = r[i];
    for ( int i = 1; i <= n; i ++ )
        printf ( "%.2lf ", ans[i] );
    return 0;
}

 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[100005];

LL TR[400005], tag[400005], vc[100005];

void update ( int nd ) {
    TR[nd] = TR[nd << 1] + TR[nd << 1 | 1];
}

void push_down ( int nd, int l, int r ) {
    if ( tag[nd] ) {
        int mid = ( l + r ) >> 1;
        TR[nd << 1] += tag[nd] * ( mid - l + 1 );
        TR[nd << 1 | 1] += tag[nd] * ( r - mid );
        tag[nd << 1] += tag[nd];
        tag[nd << 1 | 1] += tag[nd];
        tag[nd] = 0;
    }
}

void build ( int nd, int l, int r ) {
    TR[nd] = 0; tag[nd] = 0;
    if ( l == r ) {
        TR[nd] = a[vc[l]];
        return ;
    }
    int mid = ( l + r ) >> 1;
    build ( nd << 1, l, mid );
    build ( nd << 1 | 1, mid + 1, 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 + 1 ) * d;
        tag[nd] += d;
        return ;
    }
    push_down ( nd, l, r );
    int mid = ( l + r ) >> 1;
    if ( L <= mid ) add ( nd << 1, l, mid, L, R, d );
    if ( R > mid ) add ( nd << 1 | 1, mid + 1, 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 ) >> 1;
    if ( pos <= mid ) return query ( nd << 1, l, mid, pos );
    else return query ( nd << 1 | 1, mid + 1, r, pos );
}

bool check ( LL mid ) {
    int tot = 0; LL sum = 0;
    for ( int i = 1; i <= n; i ++ )
        if ( a[i] < mid ) vc[++tot] = i;
    build ( 1, 1, tot );
    vc[++tot] = 0x7f7f7f7f7f7f7f;
    for ( int i = 1; i < tot; i ++ ) {
        LL now = query ( 1, 1, tot - 1, i );
        if ( now >= mid ) continue;
        if ( mid - now + sum > T ) { sum = T + 1; break; }
        int to = vc[i] + K - 1;
        int pos = upper_bound ( vc + 1, vc + 1 + tot, to ) - vc - 1;
        add ( 1, 1, tot - 1, i, pos, mid - now );
        sum += mid - now;
    }
    if ( sum <= T ) return 1;
    return 0;
}

LL MI = 0x3f3f3f3f, MA;
LL erfen ( ) {
    LL l = MI, r = MA + T, ans;
    while ( l <= r ) {
        int mid = ( l + r ) >> 1;
        if ( check ( mid ) ) l = mid + 1, ans = mid;
        else r = mid - 1;
    }
    return ans;
}

int main ( ) {
    freopen ( "watering.in", "r", stdin );
    freopen ( "watering.out", "w", stdout );
    scanf ( "%d%d%I64d", &n, &K, &T );
    for ( int i = 1; i <= n; i ++ )    scanf ( "%I64d", &a[i] ), MI = min ( MI, a[i] ), MA = max ( MA, a[i] );
    LL ans = erfen ( );
    printf ( "%I64d", ans );
    return 0;
}