hdu5141——LIS again

最新推荐文章于 2024-07-31 10:00:00 发布

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最新推荐文章于 2024-07-31 10:00:00 发布

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分类专栏： DP hdu 数据结构文章标签： dp 线段树

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本文介绍了一种结合线段树与动态规划（DP）的高效算法，用于解决特定类型的最长递增子序列（LIS）问题。通过实例演示了如何利用线段树维护最长递增子序列的相关信息，并在O(n)时间内统计所有符合条件的子区间。

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LIS again

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 132 Accepted Submission(s): 34

Problem Description

A numeric sequence of ai is ordered if

a1<a2<…<aN . Let the subsequence of the given numeric sequence (

a1,a2,…,aN ) be any sequence (

ai1,ai2,…,aiK ), where

1≤i1<i2<…<iK≤N . For example, sequence (1, 7, 3, 5, 9, 4, 8) has ordered subsequences, eg. (1, 7), (3, 4, 8) and many others.
S[ i , j ] indicates (

ai,ai+1,ai+2,…,aj ) .
Your program, when given the numeric sequence (

a1,a2,…,aN ), must find the number of pair ( i, j) which makes the length of the longest ordered subsequence of S[ i , j ] equals to the length of the longest ordered subsequence of (

a1,a2,…,aN ).

Input

Multi test cases (about 100), every case occupies two lines, the first line contain n, then second line contain n numbers

a1,a2,…,aN separated by exact one space.
Process to the end of file.

[Technical Specification]

1≤n≤100000

0≤ai≤1000000000

Output

For each case，.output the answer in a single line.

Sample Input

Sample Output

1
3

Source

BestCoder Round #21

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线段树+dp，dp[i]表示以第i个元素结尾的LIS的长度，线段树维护此LIS的最靠右的起始位置和长度，靠右是是为了完全统计出这样的区间，最后只要O(n)的时间去统计

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 100010;
const int inf = -0x3f3f3f3f;
int xis[N];
int arr[N];
int dp[N];
int sta[N];
int end[N];
int ans, s;
int cnt;

struct node
{
	int l, r;
	int val, s;
}tree[N << 2];

void build(int p, int l, int r)
{
	tree[p].l = l;
	tree[p].r = r;
	tree[p].val = inf;
	tree[p].s = inf;
	if (l == r)
	{
		return;
	}
	int mid = (l + r) >>1;
	build(p << 1, l, mid);
	build(p << 1 | 1, mid + 1, r);
}

int BinSearch(int val)
{
	int l = 1, r = cnt, mid;
	int ans;
	while (l <= r)
	{
		mid = (l + r) >> 1;
		if (xis[mid] > val)
		{
			r = mid - 1;
		}
		else if (xis[mid] < val)
		{
			l = mid + 1;
		}
		else
		{
			ans = mid;
			break;
		}
	}
	return ans;
}

void update(int p, int pos, int val, int s)
{
	if (tree[p].l == tree[p].r)
	{
		if (tree[p].val == val && tree[p].s < s)
		{
			tree[p].s = s;
		}
		else if (tree[p].val < val)
		{
			tree[p].val = val;
			tree[p].s = s;
		}
		return;
	}
	int mid = (tree[p].l + tree[p].r) >> 1;
	if (pos <= mid)
	{
		update(p << 1, pos, val, s);
	}
	else
	{
		update(p << 1 | 1, pos, val, s);
	}
	if (tree[p << 1].val > tree[p << 1 | 1].val)
	{
		tree[p].val = tree[p << 1].val;
		tree[p].s = tree[p << 1].s;
	}
	else if (tree[p << 1].val < tree[p << 1 | 1].val)
	{
		tree[p].val = tree[p << 1 | 1].val;
		tree[p].s = tree[p << 1 | 1].s;
	}
	else
	{
		tree[p].s = max(tree[p << 1].s, tree[p << 1 | 1].s);
	}
}

void query(int p, int l, int r)
{
	if (tree[p].l >= l && tree[p].r <= r)
	{
		if (tree[p].val > ans)
		{
			ans = tree[p].val;
			s = tree[p].s;
		}
		else if (tree[p].val == ans && s < tree[p].s)
		{
			s = tree[p].s;
		}
		return;
	}
	int mid = (tree[p].l + tree[p].r) >> 1;
	if (r <= mid)
	{
		query(p << 1, l, r);
	}
	else if (l > mid)
	{
		query(p << 1 | 1, l, r);
	}
	else
	{
		query(p << 1, l, mid);
		query(p << 1 | 1, mid + 1, r);
	}
}

int main()
{
	int n;
	while (~scanf("%d", &n))
	{
		__int64 ret = 0;
		cnt = 0;
		int tmp = 0;
		for (int i = 1; i <= n; ++i)
		{
			scanf("%d", &arr[i]);
			xis[++cnt] = arr[i];
		}   
		if (n == 1)
		{
			printf("1\n");
			continue;
		}
		sort (xis + 1, xis + cnt + 1);
		cnt = unique(xis + 1, xis + cnt + 1) - xis - 1;
		dp[1] = 1;
		sta[1] = 1;
		build(1, 1, cnt);
		update(1, BinSearch(arr[1]), dp[1], sta[1]);
		for (int i = 2; i <= n; ++i)
		{
			ans = inf;
			s = inf;
			if (BinSearch(arr[i]) == 1)
			{
				dp[i] = 1;
				sta[i] = i;
			}
			else
			{
				query(1, 1, BinSearch(arr[i]) - 1);
			}
			if (ans == inf)
			{
				dp[i] = 1;
				sta[i] = i;
			}
			else
			{
				dp[i] = ans + 1;
				sta[i] = s;
			}
			update(1, BinSearch(arr[i]), dp[i], sta[i]);
			tmp = max(tmp, dp[i]);
		}
		int last = n + 1;
		for (int i = n; i >= 1; --i)
		{
			if (dp[i] == tmp)
			{
				end[i] = last - 1;
				last = i;
			}
		}
		for (int i = 1; i <= n; ++i)
		{
			if (dp[i] == tmp)
			{
				ret += (__int64)sta[i] * (__int64)(end[i] - i + 1);
			}
		}
		printf("%I64d\n", ret);
	}
	return 0;
}