Codeforces Round #248 (Div. 2) C. Ryouko's Memory Note
2014-05-30 16:43
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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。
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; }
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