hdu1025 - LIS_javahdu1025-优快云博客

本文链接：https://blog.youkuaiyun.com/tlight/article/details/51703251

本文介绍了一种求解最长递增子序列(LIS)问题的高效算法，并提供了两种实现方式：一种是时间复杂度为O(nlogn)的方法，通过动态规划结合二分查找来优化；另一种是传统的时间复杂度为O(n^2)的动态规划方法。通过对具体代码的解析，展示了如何在实际应用中计算城市间道路建设的最大递增序列长度。

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// LIS
// O(nlogn)

#include <stdio.h>

#define MAX_N 500005

int n;			// number of roads plan to build
int road[MAX_N]; // road[p]=r, means a road from city p to city r, city start from 1
int dp[MAX_N];	// dp[i] means the length of LIS from 1 to i
int b[MAX_N];	// b[j] means the minimum value of the last element of the LIS whose length is j

int max(int a, int b)
{
	return a > b ? a : b;
}

int binary_search(int a[], int size, int value)
{
	int lo = 1;
	int hi = size;
	while (lo < hi)
	{
		int mid = (lo + hi) / 2;
		if (a[mid] < value)
		{
			lo = mid + 1;
		}
		else
		{
			hi = mid;
		}
	}
	return lo;
}

int lis()
{
	int i = 1;
	int sz = 1;
	b[1] = road[i];
	dp[1] = 1;
	for (i = 2; i <= n; i++)
	{
		if (road[i] < b[1])
		{
			b[1] = road[i];
			dp[i] = 1;
		}
		else if (road[i] > b[sz])
		{
			sz++;
			b[sz] = road[i];
			dp[i] = sz;
		}
		else
		{	// find the k, which b[k-1]<road[i]<b[k]
			int k = binary_search(b, sz, road[i]);
			b[k] = road[i];
			dp[i] = k;
		}
	}

	int ret = 0;
	for (i = 1; i <= n; i++)
	{
		if (ret < dp[i])
			ret = dp[i];
	}
	return ret;
}


// O(n^2)
int dp_lis()
{
	int i, j;
	dp[1] = 1;
	for (i = 2; i <= n; i++)
	{
		dp[i] = 1;
		for (j = 1; j < i; j++)
		{
			if (road[i] > road[j])
			{
				if (dp[i] < dp[j] + 1)
					dp[i] = dp[j] + 1;
			}
		}
	}

	int ret = 0;
	for (i = 1; i <= n; i++)
	{
		if (ret < dp[i])
			ret = dp[i];
	}
	return ret;
}
int main()
{
	freopen("hdu1025_input.txt", "r", stdin);
	int tc = 0;
	int i;
	int p, r; // city start from 1
	int ans;	// at least can build one road
	while (scanf("%d", &n) != EOF)
	{
		tc++;
		for (i = 0; i < n; i++)
		{
			scanf("%d %d", &p, &r);
			road[p] = r;
		}

		//ans = lis();
		ans = dp_lis();

		if (ans == 1)
			printf("Case %d:\nMy king, at most %d road can be built.\n\n", tc, ans);
		else
			printf("Case %d:\nMy king, at most %d roads can be built.\n\n", tc, ans);
	}
	return 0;
}