AtCoder Regular Contest 120 C - Swaps 2(树状数组)

题目链接：点我点我

Score : $500$ points

Problem Statement

Given are two sequences of length $N$ each: $A=(A_1,A_2,A_3,\dots ,A_N)$ and $B=(B_1,B_2,B_3,\dots ,B_N)$.
Determine whether it is possible to make $A$ equal $B$ by repeatedly doing the operation below (possibly zero times). If it is possible, find the minimum number of operations required to do so.

• Choose an integer $i$ such that $1\le i<N$, and do the following in order:
  • swap $A_i$ and $A_{i+1}$;
  • add $1$ to $A_i$;
  • subtract $1$ from $A_{i+1}$.

Constraints

• $2\le N\le 2\times {10}^5$
• $0\le A_i\le {10}^9$
• $0\le B_i\le {10}^9$
• All values in input are integers.

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Input

Input is given from Standard Input in the following format:

$N$
$A_1$ $A_2$ $A_3$ $\dots$ $A_N$
$B_1$ $B_2$ $B_3$ $\dots$ $B_N$

Output

If it is impossible to make $A$ equal $B$, print -1.
Otherwise, print the minimum number of operations required to do so.

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Sample Input 1

3
3 1 4
6 2 0

Sample Output 1

2

We can match $A$ with $B$ in two operations, as follows:

• First, do the operation with $i=2$, making $A=(3,5,0)$.
• Next, do the operation with $i=1$, making $A=(6,2,0)$.

We cannot meet our objective in one or fewer operations.

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Sample Input 2

3
1 1 1
1 1 2

Sample Output 2

-1

In this case, it is impossible to match $A$ with $B$.

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Sample Input 3

5
5 4 1 3 2
5 4 1 3 2

Sample Output 3

0

$A$ may equal $B$ before doing any operation.

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Sample Input 4

6
8 5 4 7 4 5
10 5 6 7 4 1

Sample Output 4

7

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给出两个数组，每个数组有 n 个数，可以进行如下操作 ：
将 a 数组中的 a[i]->a[i]+1 与 a[i+1]->a[i+1]-1 然后再交换位置，问最少需要多少次操作可以将 a 数组变为 b 数组

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先看一下第 4 组样例该怎么做，

\[8,5,4,7,4,5 \]

想让第一个数变为 10，我们可以让 7 移动到第一个位置上来，也可以让最后一个 5 移动到第一个位置上来，但是就这一步而言让 7 过来是最优的，而 7 前面的 8 5 4，分别要减去 1；
所以现在的思路有了，对于 b 数组中的每一个位置 j，我们只要找到 a 数组中 \(a[i]+i-j=b[j]\) 的第一个 \(a[i]\) 即可，时间复杂度 \(O(n^2)\)

这样是远远不够的，我们需要更快速地方法，我们将上式做一个变化

\[a[i]+i=b[j]+j \]

这个式子就很漂亮，这样将 a 和 b 数组同时变为 \((a[i]+i)\) ，
\((b[i]+i)\) 的形式，这样 a ,b 数组就分别变为了

\[9 ,7 ,7 ,11 ,9 ,11 \]

\[11,7,9,11,9,7 \]

这个问题，两个数组相同，求交换多少次可以变成另一个数组，这不就是求 ‘逆序对’ 吗

const int N = 3e5 + 5;

    ll n, m, _;
    int i, j, k;
    int a[N];
    int b[N], c[N];
    map<int, vector<int>> mp;

void add(int x)
{
    for(; x <= n; x += lowbit(x)){
        c[x] ++;
    }
}

int ask(int x)
{
    int ans = 0;
    for(; x; x -= lowbit(x)){
        ans += c[x];
    }
    return ans;
}

signed main()
{
    //IOS;
    while(~ sd(n)){
        rep(i, 1, n) sd(a[i]);
        rep(i, 1, n) sd(b[i]);
        ll ok = 0;
        per(i, 1, n){
            a[i] += i;
            b[i] += i;
            mp[b[i]].pb(i);
            ok += a[i] - b[i];
        }
        rep(i, 1, n){
            int tmp = a[i];
            if(mp[a[i]].empty()){
                ok = 1;
                break;
            }
            a[i] = *mp[a[i]].rbegin();
            mp[tmp].pop_back();
        }
        if(ok){
            puts("-1");
            break;
        }
        ll ans = 0;
        per(i, 1, n){
            ans += ask(a[i]);
            add(a[i]);
        }
        pll(ans);
    }
    return 0;
}