POI2008BBB-BBB

Year2008 #POI #贪心 #数学

考虑枚举旋转了几次，维护一个前缀和，一个前缀和的 \(min\) ，目标为使和合法并使前缀 \(min\) 满足条件，这个代价就可以 \(\mathcal{O}(1)\) 计算

// Author: xiaruize
const int INF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1000000007;
const int N = 2e6 + 10;

int n, p, q, x, y;
char s[N];
int a[N], mi[N];
int dp[N];

void solve()
{
	cin >> n >> p >> q >> x >> y;
	cin >> (s + 1);
	rep(i, 1, n)
	{
		if (s[i] == '-')
			a[i] = -1;
		else
			a[i] = 1;
		a[i] += a[i - 1];
	}
	rep(i, 1, n) mi[i] = min(mi[i - 1], a[i]);
	int d = (q - p - a[n]) / 2;
	int res = INF;
	rep(i, 1, n)
	{
		int cur = max(0, (-p - min(a[n] - a[n - i + 1] + mi[n - i + 1], dp[i - 1]) + 1) / 2);
		res = min(res, (i - 1) * y + (cur + abs(d - cur)) * x);
		dp[i] = min(dp[i - 1] + a[n - i + 1] - a[n - i], 0);
	}
	cout << res << endl;
}

#ifndef ONLINE_JUDGE
bool end_of_memory_use;
#endif

signed main()
{
	// freopen(".in","r",stdin);
	// freopen(".out","w",stdout);
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int testcase = 1;
	// cin >> testcase;
	while (testcase--)
		solve();
#ifndef ONLINE_JUDGE
	cerr << "Memory use:" << (&end_of_memory_use - &start_of_memory_use) / 1024.0 / 1024.0 << "MiB" << endl;
	cerr << "Time use:" << (double)clock() / CLOCKS_PER_SEC * 1000.0 << "ms" << endl;
#endif
	return 0;
}