标签:acm codeforces
const int maxn = 110000; LL ipt[maxn]; map<LL, vector<LL> > mp; map<LL, vector<LL> >::iterator it; int main() { int n, m; while (~RII(n, m)) { mp.clear(); REP(i, m) { cin >> ipt[i]; } REP(i, m) { if (i > 0 && ipt[i - 1] != ipt[i]) mp[ipt[i]].push_back(ipt[i - 1]); if (i < m - 1 && ipt[i + 1] != ipt[i]) mp[ipt[i]].push_back(ipt[i + 1]); } LL ans = 0; FC(it, mp) { vector<LL>& ipt = it->second; sort(all(ipt)); LL t = 0, m = ipt[(LL)ipt.size() / 2]; REP(i, ipt.size()) { t += abs((it->first) - ipt[i]) - abs(m - ipt[i]); } ans = max(ans, t); } ans *= -1; REP(i, m - 1) ans += abs(ipt[i] - ipt[i + 1]); cout << ans << endl; } return 0; }
Codeforces Round #248 (Div. 1)——Ryouko's Memory Note,布布扣,bubuko.com
Codeforces Round #248 (Div. 1)——Ryouko's Memory Note
标签:acm codeforces
原文地址:http://blog.csdn.net/wty__/article/details/37902781