HDU 6070 (线段树)(统计颜色)

HDU 6070 Partition

Problem :

给一段长度为n的序列，要求找出一段区间，使得这段区间的数字种类除以区间长度最小。输出最后的答案即可。(n <= 60000)(9s时限)

Solution :

显然，答案是0~1中的一个数字，可以很自然的想到二分答案的做法。假设目前二分到的答案为mid，那么需要判断

\[\frac{cnt(l, r)}{r- l + 1} <= mid
\]

其中cnt(l,r)为l到r这个区间内的数字种类。变化一下式子可以得到:

\[cnt(l, r) + mid * l <= mid * (r + 1)
\]

通过枚举有端点r，使用线段树维护左边的式子，每当右端点r向右移动1时，所影响的区间为r到对应颜色上一次出现的位置，区间整体加1就行了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <string>
#include <cmath>
#include <algorithm>
#include <vector>
#include <map>
#include <queue>

using namespace std;

#define eps 1e-10

const int N = 1e5 + 8;

int a[N], pre[N];
int n;

struct Segment_Tree
{
	double tag[N << 2];
	double lazy[N << 2];
	void pushup(int rt)
	{
		int l = rt << 1, r = rt << 1 | 1;
		tag[rt] = min(tag[l], tag[r]);
	}
	void pushdown(int rt)
	{
		int l = rt << 1, r = rt << 1 | 1;
		if (lazy[rt])
		{
			tag[l] += lazy[rt];
			tag[r] += lazy[rt];
			lazy[l] += lazy[rt];
			lazy[r] += lazy[rt];
			lazy[rt] = 0;
		}
	}
	void build(int l, int r, int rt, double x)
	{
		tag[rt] = lazy[rt] = 0;
		if (l == r)
		{
			tag[rt] = l * x;
			return;
		}
		int m = l + r >> 1;
		build(l, m, rt << 1, x);
		build(m + 1, r, rt << 1 | 1, x);
		pushup(rt);
	}
	void update(int L, int R, int val, int l, int r, int rt)
	{
		if (L <= l && r <= R)
		{
			tag[rt] = tag[rt] + val;
			lazy[rt] += val;
			return;
		}
		pushdown(rt);
		int m = l + r >> 1;
		if (L <= m) update(L, R, val, l, m, rt << 1);
		if (m <  R) update(L, R, val, m + 1, r, rt << 1 | 1);
		pushup(rt);
	}
	double query(int L, int R, int l, int r, int rt)
	{
		if (L <= l && r <= R)
		{
			return tag[rt];
		}
		pushdown(rt);
		int m = l + r >> 1;
		double ans = 1e12;
		if (L <= m) ans = min(ans, query(L, R, l, m, rt << 1));
		if (m <  R) ans = min(ans, query(L, R, m + 1, r, rt << 1 | 1));
		return ans;
	}
}T;

void init()
{
	cin >> n;
	for (int i = 1; i <= n; ++i) cin >> a[i];
}
int sgn(double x)
{
	if (fabs(x) < eps) return 0;
	if (x > 0) return 1;
	return -1;
}

bool check(double mid)
{

	for (int i = 1; i <= n; ++i) pre[i] = 0;
	T.build(1, n, 1, mid);
	for (int i = 1; i <= n; ++i)
	{
		T.update(pre[a[i]] + 1, i, 1, 1, n, 1);
		pre[a[i]] = i;
		if (sgn(mid * (i + 1) - T.query(1, i, 1, n, 1) >= 0)) return 1;
	}
	return 0;
}

void solve()
{
	double l = 0, r = 1;
	while (l + eps < r)
	{
		double mid = (l + r) / 2;
		if (check(mid)) r = mid; else l = mid;
	}
	printf("%.6f\n", l);

}

int main()
{
	cin.sync_with_stdio(0);
    int T; cin >> T;
    for (int cas = 1; cas <= T; ++cas)
    {
        init();
        solve();
    }
}