Simplified Reversi （非线段树法，图解）————真的抽象

更逊的体验

1.前言

这道题不用线段树是真的抽象。。。

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2.题解

我们只分析在列上放棋子

0.变量

以右下角为(1,1)
$Min1$ ：最大的修改行上的$x$
$Min2$ ：最大的修改列上的$x$
$Minl[i]$ ：第$i$列的最大的修改行的x
$Minr[i]$ ：第$i$行的最大的修改列的x

1. 若 $x\in [1,Min1]$

如图
则粉色那条线的$Minl$值不会再被更新
所以$Minl[i]=Min2(i\in [x,Min1])$ (更新)
$val+=Minl[x]$
注：灰色那条线是$x$

2. $x\in (Min1,n-2]$

如图
则灰色这条线不会修改任意的$Minl$
$val+=Minl[x]$
注：灰色那条线是$x$

3.代码

#include <vector>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

#define LL long long

template <typename T> void read (T &x) {x = 0; T f = 1;char tem = getchar ();while (tem < '0' || tem > '9') {if (tem == '-') f = -1;tem = getchar ();}while (tem >= '0' && tem <= '9') {x = (x << 1) + (x << 3) + tem - '0';tem = getchar ();}x *= f;}
template <typename T> void write (T x) {if (x < 0) {x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0');}
template <typename T> T Max (T x, T y) { return x > y ? x : y; }
template <typename T> T Min (T x, T y) { return x < y ? x : y; }
template <typename T> T Abs (T x) { return x > 0 ? x : -x; }

const int Maxn = 2 * 1e5 + 5;

LL n, q, ans;
LL Min1, Min2;
LL Minl[Maxn], Minr[Maxn];

int main () {
	read (n); read (q);
	Min1 = Min2 = n;
	for (int i = 1; i <= n; i++) Minl[i] = Minr[i] = n;
	for (int i = 1; i <= q; i++) {
		LL type, x;
		read (type); read (x);
		if (type == 1) {
			if (x < Min1) {//放在[1, Min1]中 
				for (int i = x; i <= Min1; i++) {
					Minl[i] = Min (Minl[i], Min2);
				}
			}
			ans += Minl[x] - 2;
			Min1 = Min (Min1, x);
		}
		else {
			if (x < Min2) {
				for (int i = x; i <= Min2; i++) {
					Minr[i] = Min (Minr[i], Min1);
				}
			}
			ans += Minr[x] - 2;
			Min2 = Min (Min2, x);
		}
	}
	write ((n - 2) * (n - 2) - ans);
    return 0;
}