CF Little Elephant and Shifts

最新推荐文章于 2020-02-21 03:04:09 发布

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最新推荐文章于 2020-02-21 03:04:09 发布

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本文详细阐述了如何计算两个排列之间的最小距离，并通过具体例子解释了循环移位的概念及其应用。利用标准集合数据结构，提供了解决问题的高效算法，包括如何在循环移位过程中更新距离值。

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题目：http://blog.youkuaiyun.com/wukonwukon/article/details/7945526

代码：http://blog.youkuaiyun.com/acm_ted/article/details/7943188

官方题解：http://codeforces.com/blog/entry/5177

给定两个两个1-n的排列，定义两个排列的距离为：所有相同数字之间距离的最小值，More formally, it's such minimum |i - j|, that a_i = b_j.定义A cyclic shift number i (1 ≤ i ≤ n) of permutation b consisting from n elements is a permutation b_ib_i + 1... b_nb₁b₂... b_i - 1. Overall a permutation has n cyclic shifts. 输出者n个cyclic shifts的距离。

eg:

4
2 1 3 4
3 4 2 1

输出 2 1 0 1

220C - Little Elephant and Shifts

Each of the shifts can be divided into two parts — the right (the one that starts from occurrence 1) and the left (the rest of the elements). If we could keep minimal distance for each part, the minimal of these numbers will be the answers for the corresponding shift. Lets solve the problems of the right part, the left will be almost the same.

Let we have some shift, for example 34567[12] and the permutation A is 4312765 and B is 2145673, then shifted B is 4567321. Let we keep two sets (S1 and S2). The first will keep all the distances from integers in current left part to the corresponding positions in A(for the example above, it is \texttt{2, 4}). When you come to the next shift, all integers in S1 should be decreased by 1 (that is because all distances are also decreased by 1). But now some integers in set may be negative, when any negative integer occures (it always will be -1) you need to delete it from S1 and put 1 to the S2. Also after shifting to the next shifts, all integers in S2 must be increase by 1. After that, for any shift, the answer will be minimum from the smallest numbers in S1 and S2.

It was very useful to use standart "set" in C++.

#include<cstdio>
#include<algorithm>
#include<set>
using namespace std;
#define MAX 100010
int a[MAX],b[MAX],hashx[MAX],d[MAX];
multiset<int> s;
int main()
{
    int n,t,res;
    scanf("%d",&n);
    for(int i=0; i<n; ++i)
    {
        scanf("%d",a+i);
        hashx[a[i]]=i;//存放数字在a中的位置
    }
    for(int i=0; i<n; ++i)
    {
        scanf("%d",b+i);
        s.insert(i-hashx[b[i]]);//插入相同数字的距离
    }
    for(int i=0; i<n; ++i)
    {
        multiset<int>::iterator it=s.lower_bound(i);//查找大于等于i的第一个数
        res=0xfffff;
        if(it!=s.end())   res=min(res,(*it)-i);//非负数中最小的数的绝对值
        if(it!=s.begin()) res=min(res,i-(*(--it)));//负数中的最大的数的绝对值
        printf("%d\n",res);
        //模拟旋转
        t=b[i];
        s.erase(s.find(i-hashx[t]));
        s.insert((i+n)-hashx[t]);
    }
    return 0;
}