Solution -「ARC 063D」「AT 2149」Snuke's Coloring 2
\(\mathcal{Decription}\)
Link.
平面上有一个左下角坐标 \((0,0)\) 右上角坐标 \((W,H)\) 的矩形,起初长方形内部被涂白。
现在给定 \(n\) 个点,你每次在以下 \(4\) 种操作中选择一种:
- 将矩形内 \(x<x_i\) 的区域涂黑;
- 将矩形内 \(x>x_i\) 的区域涂黑;
- 将矩形内 \(y<y_i\) 的区域涂黑;
- 将矩形内 \(y>y_i\) 的区域涂黑。
最大化操作后白色矩阵周长。
\(n\le3\times10^5\),\(W,H\le10^8\)。
\(\mathcal{Solution}\)
就挺 amazing 的题呐。
题意等价于求周长最大的矩形,使得矩形内不包含任意一个点。
首先,答案有下界 \(2\max\{H,W\}+2\)。考虑一个周长超过该下界的矩形,它一定跨过 \(y=\frac{H}2\) 或 \(x=\frac{W}2\),所以只需要分别求出跨过着两条直线的周长最大的合法矩形。下以跨过 \(l:y=\frac{H}2\) 的情形为例。
从左到右用一条扫描线,设当前 \(x=x_0\) 轴上高于 \(l\) 的最低点离 \(l\) 的距离为 \(u\),低于 \(l\) 的最高点离 \(l\) 的距离为 \(d\),那么当矩形过 \(x=x_0\) 时,高度 \(\le u+d\)。
接下来,把从 \(x_1\) 到 \(x_2\) 的横向宽度表达为 \((W-x_1)-(W-x_2)\),然后线段树维护矩形左边界为 \(x=x_0\) 时,高度 \(+(W-x_1)\) 的值。用单调递减的单调栈维护 \(u\) 和 \(d\),弹栈时修改一段区间的值,最后求前缀最大值即可。(建议参照代码理解。)
\(\mathcal{Code}\)
/* Clearink */
#include <stack>
#include <cstdio>
#include <algorithm>
inline int rint () {
int x = 0; char s = getchar ();
for ( ; s < '0' || '9' < s; s = getchar () );
for ( ; '0' <= s && s <= '9'; s = getchar () ) x = x * 10 + ( s ^ '0' );
return x;
}
inline int min_ ( const int a, const int b ) { return a < b ? a : b; }
inline int max_ ( const int a, const int b ) { return a < b ? b : a; }
const int MAXN = 3e5;
int n, W, H, x[MAXN + 5], up[MAXN + 5], dn[MAXN + 5];
std::stack<int> stku, stkd;
struct Point {
int x, y;
inline bool operator < ( const Point& p ) const { return x < p.x; }
} p[MAXN + 5];
struct SegmentTree {
int mx[MAXN << 2], tag[MAXN << 2];
inline void pushup ( const int rt ) {
mx[rt] = max_ ( mx[rt << 1], mx[rt << 1 | 1] );
}
inline void pushdn ( const int rt ) {
int& k = tag[rt];
if ( !k ) return ;
mx[rt << 1] += k, tag[rt << 1] += k;
mx[rt << 1 | 1] += k, tag[rt << 1 | 1] += k;
k = 0;
}
inline void clear ( const int rt, const int l, const int r ) {
mx[rt] = tag[rt] = 0;
if ( l == r ) return ;
int mid = l + r >> 1;
clear ( rt << 1, l, mid ), clear ( rt << 1 | 1, mid + 1, r );
}
inline void update ( const int rt, const int l, const int r,
const int ul, const int ur, const int uv ) {
if ( ul <= l && r <= ur ) return mx[rt] += uv, tag[rt] += uv, void ();
int mid = l + r >> 1; pushdn ( rt );
if ( ul <= mid ) update ( rt << 1, l, mid, ul, ur, uv );
if ( mid < ur ) update ( rt << 1 | 1, mid + 1, r, ul, ur, uv );
pushup ( rt );
}
inline int query ( const int rt, const int l, const int r,
const int ql, const int qr ) {
if ( ql <= l && r <= qr ) return mx[rt];
int mid = l + r >> 1, ret = 0; pushdn ( rt );
if ( ql <= mid ) ret = query ( rt << 1, l, mid, ql, qr );
if ( mid < qr ) ret = max_ ( ret, query ( rt << 1 | 1, mid + 1, r, ql, qr ) );
return ret;
}
} segt;
inline int solve ( const int n, const int W, const int H ) {
if ( !n ) return 0;
int mid = H >> 1, cnt = 0, ret = 0;
for ( ; !stku.empty (); stku.pop () );
for ( ; !stkd.empty (); stkd.pop () );
segt.clear ( 1, 1, n );
stku.push ( 0 ), stkd.push ( 0 );
for ( int i = 1; i <= n; ) {
x[++ cnt] = p[i].x;
up[cnt] = H - mid, dn[cnt] = mid;
for ( ; i <= n && p[i].x == x[cnt]; ++ i ) {
if ( p[i].y >= mid ) up[cnt] = min_ ( up[cnt], p[i].y - mid );
if ( p[i].y <= mid ) dn[cnt] = min_ ( dn[cnt], mid - p[i].y );
}
int las;
while ( up[las = stku.top ()] > up[cnt] ) {
stku.pop ();
segt.update ( 1, 1, n, stku.top () + 1, las, up[cnt] - up[las] );
}
stku.push ( cnt );
while ( dn[las = stkd.top ()] > dn[cnt] ) {
stkd.pop ();
segt.update ( 1, 1, n, stkd.top () + 1, las, dn[cnt] - dn[las] );
}
stkd.push ( cnt );
segt.update ( 1, 1, n, cnt, cnt, W - x[cnt - 1] + up[cnt] + dn[cnt] );
ret = max_ ( ret, segt.query ( 1, 1, n, 1, cnt ) - W + p[i].x );
}
return ret;
}
int main () {
W = rint (), H = rint (), n = rint ();
for ( int i = 1; i <= n; ++ i ) {
p[i].x = rint (), p[i].y = rint ();
if ( !p[i].x || p[i].x == W || !p[i].y || p[i].y == H ) -- i, -- n;
}
int ans = 0;
std::sort ( p + 1, p + n + 1 ), p[n + 1].x = W;
for ( int i = 1; i <= n + 1; ++ i ) {
ans = max_ ( ans, H + p[i].x - p[i - 1].x );
}
ans = max_ ( ans, solve ( n, W, H ) );
for ( int i = 1; i <= n; ++ i ) p[i].x ^= p[i].y ^= p[i].x ^= p[i].y;
std::sort ( p + 1, p + n + 1 ), p[n + 1].x = H;
for ( int i = 1; i <= n + 1; ++ i ) {
ans = max_ ( ans, W + p[i].x - p[i - 1].x );
}
ans = max_ ( ans, solve ( n, H, W ) );
printf ( "%d\n", ans << 1 );
return 0;
}