Codeforces Round #248 (Div. 2) C. Ryouko's Memory Note

 题目链接：http://codeforces.com/contest/433/problem/C

思路：可以想到，要把某一个数字变成他的相邻中的数字的其中一个，这样总和才会减少，于是我们可以把每个数的左右两个相邻的数字存起来，然后我们可以想到，把某个数变成这些相邻的数的中位数总和最小。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#define REP(i, a, b) for (int i = (a); i < (b); ++i)
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
using namespace std;

const int MAX_N = (100000 + 100);
int N, M, a[MAX_N];
long long sum, ans;
vector<int > neighbor[MAX_N];

int main()
{
while (cin >> N >> M) {
FOR(i, 1, N) neighbor[i].clear();
FOR(i, 1, M) cin >> a[i];
sum = 0;
FOR(i, 2, M) if (a[i - 1] != a[i]) {
neighbor[a[i - 1]].push_back(a[i]);
neighbor[a[i]].push_back(a[i - 1]);
sum += abs(a[i - 1] - a[i]);
}
ans = sum;
FOR(i, 1, N) if ((int)neighbor[i].size()) {
sort(neighbor[i].begin(), neighbor[i].end());
int _size = (int)neighbor[i].size();
long long tmp = sum;
int target = neighbor[i][_size / 2];
REP(j, 0, _size) {
tmp = tmp - abs(i - neighbor[i][j]) + abs(target - neighbor[i][j]);
}
ans = min(ans, tmp);
}
cout << ans << endl;
}
return 0;
}