HDU 1950 Bridging signals 解题报告

HDU 1950 Bridging signals 解题报告

解题思路：这题目看起来烦，实际上就是让你找最长上升子序列，我一开始还被唬住了。。。有两种找的方法，一种是dp算法，一种是LIS算法。
1.dp方法。一个不幸的消息，tle了，这道题上dp算法确实慢点。

#include<iostream>
#include<math.h>
#include<iomanip>
#include<algorithm>
#include<iostream>
#include<math.h>
#include<iomanip>
#include<algorithm>
#include<queue>
#include<cstring>
#include<string>
#include<map>
#include<stack>
#include<stdio.h>
#include<cstdio>
#include<stdlib.h>
#include<fstream>
#include<iomanip>
#pragma warning(disable:4996)
#define INF 0x3f3f3f3f
#define ll long long
#define PI acos(-1.0)
const int N = 1000010;
const int maxn = 1e9;
using namespace std;
int n,p;
int a[40005];
int dp[40005];
int main()
{
	cin >> n;
	while (n--)
	{
		int ans=-1;
		memset(dp, 0, sizeof(dp));
		cin >> p;
		for(int i = 1; i <= p; i++)
		{			
			cin >> a[i];
			dp[i] = 1;
		}
		for (int i = 2; i <= p; i++)
		{
			for (int j = 1; j <= i; j++)
			{
				if (a[i] > a[j])
				{
					dp[i] = max(dp[i], dp[j] + 1);
				}
			}
			ans = max(ans, dp[i]);
		}		
		cout << ans<< endl;
	}
}

2.LIS算法。发个链接吧：https://blog.csdn.net/Fire_to_cheat_/article/details/78297534。我懒的讲思维了。如果还看不懂就举例子想，多想想总会懂的。

#include<iostream>
#include<math.h>
#include<iomanip>
#include<algorithm>
#include<iostream>
#include<math.h>
#include<iomanip>
#include<algorithm>
#include<queue>
#include<cstring>
#include<string>
#include<map>
#include<stack>
#include<stdio.h>
#include<cstdio>
#include<stdlib.h>
#include<fstream>
#include<iomanip>
#pragma warning(disable:4996)
#define INF 0x3f3f3f3f
#define ll long long
#define PI acos(-1.0)
const int N = 1000010;
const int maxn = 1e9;
using namespace std;
int n,p;
int a[40005];
int tem[40005];
int main()
{
	cin >> n;
	while (n--)
	{
		memset(tem, 0, sizeof(tem));
		cin >> p;
		for(int i = 1; i <= p; i++)
		{			
			cin >> a[i];
		}
		tem[1] = a[1];
		int len = 1;//子序列长度最少是1
		for (int i = 2; i <= p; i++)
		{
			if (a[i] > tem[len])
			{
				len++;
				tem[len] = a[i];
			}
			else
			{
				int pos = lower_bound(tem + 1, tem + len + 1, a[i]) - tem;//因为lower_bound()是二分查找，所以很快
				tem[pos] = a[i];
			}
		}
		cout << len<< endl;
	}
}