问题 E: guruguru

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时间限制: 1 Sec 内存限制: 128 MB
提交: 43 解决: 18
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题目描述

Snuke is buying a lamp. The light of the lamp can be adjusted to m levels of brightness, represented by integers from 1 through m, by the two buttons on the remote control.
The first button is a "forward" button. When this button is pressed, the brightness level is increased by 1, except when the brightness level is m, in which case the brightness level becomes 1.
The second button is a "favorite" button. When this button is pressed, the brightness level becomes the favorite brightness level x, which is set when the lamp is purchased.
Snuke is thinking of setting the favorite brightness level x so that he can efficiently adjust the brightness. He is planning to change the brightness n−1 times. In the i-th change, the brightness level is changed from ai to ai+1. The initial brightness level is a1. Find the number of times Snuke needs to press the buttons when x is set to minimize this number.

Constraints
2≤n,m≤105
1≤ai≤m
ai≠ai+1
n, m and ai are integers.

输入

Input is given from Standard Input in the following format:
n m
a1 a2 … an

输出

Print the minimum number of times Snuke needs to press the buttons.

样例输入

4 6
1 5 1 4

样例输出

题目大意

两个按钮, 一个可以使计数器+1(计数器数字从1-m), 当前值为m时, 再+1就变成了1
另一个按钮储存了一个值x, 按一下就从任意值会变成x
n-1次操作, 由一个数组a[n]描述, 第i次操作: 将计数器从a[i]调到a[i+1]
将x设置为某个值, 使得所有操作需要按按钮的次数总和最小, 输出这个最小值
范围:
2≤n,m≤1052≤n,m≤105
1≤ai≤m1≤ai≤m
ai≠ai+1ai≠ai+1
n,m,ai都是整数n,m,ai都是整数

思路

设置一个数组p, p[i] := 如果x在i位置, 对于所有操作, 使用第二个按钮能够减少的操作次数(相对于只使用第一个按钮)
设l = a[i], r = a[i+1], 如果l>r, 令r = r+m, 这样就不用考虑上界的问题了
那么对于每一对l, r, 如果r-l<=1, 那么无论x在什么位置, 第二个按钮都不能减少操作次数
如果r-l>1, 那么
x在l+2位置使用第二个按钮能够减少1次操作, 在l+3位置能减少2次… 在r位置能够减少r-(l+2)+1次操作
所以p[l+2] += 1, p[l+3] += 2 ... p[r] += r-(l+2)+1
对每一对l, r如此处理, 得到最后的p数组
设all为只使用第一个按钮所需要的操作总次数
那么`最少操作次数 = all - max{p[i]+p[i+m]}, 1<=i<=m

以上就是基本思路, 如果不加其他优化, 直接写的话, 复杂度O(mn)O(mn)
能优化的地方是p[l+2] += 1, p[l+3] += 2 ... p[r] += r-(l+2)+1这就是复杂度里m的来源, 可以将它优化到O(1)O(1)

如果要将p[l]到p[r]依次加上1, 2, … r-l+1
我们可以这样: p[i] += 1, p[r+1] -= r-l+1 + 1, p[r+2] += r-l+1, 每次只更新这三个值, 最后再从头到尾p[i] += p[i-1]
具体的一组例子, 比如l=2, r=5

更新操作	复杂度	2	3	4	5	6	7
更新三个值	O(1)O(1)	1	0	0	0	-5	4
p[i]+=p[i-1]	O(n)O(n)	1	1	1	1	-4	0
p[i]+=p[i-1]	O(n)O(n)	1	2	3	4	0	0

所以最后总复杂度O(n+m)

/////////*这个博主写的太优秀，我已经没法添加一些自己理解的东西了，

只说一句：这个表格看的时候最好从下往上看，两次差分，这样好理解一些*/////////

参考自: 来源

代码

#include <bits/stdc++.h>

using namespace std;

const int MAXN = 3E5 + 100;
typedef long long ll;
int n, m, a[MAXN];
ll p[MAXN];

int main()
{
    ll all = 0;
    scanf("%d%d", &n, &m);
    for(int i=0; i<n; ++i) scanf("%d", a+i);
    for(int i=1; i<n; ++i)
    {
        int l = a[i-1], r = a[i];
        if(l>r) r += m;
        all += r-l;
        if(r-l>1)
        {
            p[l+2] += 1;
            p[r+1] -= (r-(l+2)+1) + 1;
            p[r+2] += (r-(l+2)+1);
        }
    }
    for(int i=1; i<=2*m; ++i) p[i] += p[i-1];
    for(int i=1; i<=2*m; ++i) p[i] += p[i-1];

    ll ans = -1;
    for(int i=1; i<=m; ++i) ans = max(ans, p[i]+p[i+m]);
    cout << all - ans << endl;

    return 0;
}