D - Snuke Panic (1D)

Time Limit: 2 sec / Memory Limit: 1024 MB

Score : 400 points

Problem Statement

Takahashi is trying to catch many Snuke.

There are five pits at coordinates 00, 11, 22, 33, and 44 on a number line, connected to Snuke's nest.

Now, NN Snuke will appear from the pits. It is known that the ii-th Snuke will appear from the pit at coordinate X_iXi​ at time T_iTi​, and its size is A_iAi​.

Takahashi is at coordinate 00 at time 00 and can move on the line at a speed of at most 11.
He can catch a Snuke appearing from a pit if and only if he is at the coordinate of that pit exactly when it appears.
The time it takes to catch a Snuke is negligible.

Find the maximum sum of the sizes of Snuke that Takahashi can catch by moving optimally.

Constraints

• 1 \leq N \leq 10^51≤N≤105
• 0 < T_1 < T_2 < \ldots < T_N \leq 10^50<T1​<T2​<…<TN​≤105
• 0 \leq X_i \leq 40≤Xi​≤4
• 1 \leq A_i \leq 10^91≤Ai​≤109
• All values in input are integers.

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Input

Input is given from Standard Input in the following format:

NN
T_1T1​ X_1X1​ A_1A1​
T_2T2​ X_2X2​ A_2A2​
\vdots⋮
T_NTN​ X_NXN​ A_NAN​

Output

Print the answer as an integer.

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Sample Input 1 Copy

Copy

3
1 0 100
3 3 10
5 4 1

Sample Output 1 Copy

Copy

101

The optimal strategy is as follows.

• Wait at coordinate 00 to catch the first Snuke at time 11.
• Go to coordinate 44 to catch the third Snuke at time 55.

It is impossible to catch both the first and second Snuke, so this is the best he can.

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Sample Input 2 Copy

Copy

3
1 4 1
2 4 1
3 4 1

Sample Output 2 Copy

Copy

0

Takahashi cannot catch any Snuke.

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Sample Input 3 Copy

Copy

10
1 4 602436426
2 1 623690081
3 3 262703497
4 4 628894325
5 3 450968417
6 1 161735902
7 1 707723857
8 2 802329211
9 0 317063340
10 2 125660016

Sample Output 3 Copy

Copy

2978279323

设f[i][j]表示最大值的一个集合， i表示第i秒， j表示第 j 个坑。

则有状态转移方程f[i][j] = a[i][j] + f[i - 1][j], 在原地不动；

f[i][j] = max(f[i][j], a[i][j] + f[i - 1][j - 1]), 从f[i - 1][j - 1] 得来

f[i][j] = max(f[i][j], a[i][j] + f[i - 1][j + 1]),向后退， 从f[i - 1][j + 1] 得来。

Code:

#pragma GCC optimize(1)
#pragma GCC optimize(2)
#pragma GCC optimize(3, "Ofast", "inline")
#include <iostream>
#include <stdio.h>
#include <cstring>
#include <math.h>
#include <queue>
#include <algorithm>
#include <map>
#include <set>
#include <unordered_map>

using namespace std;

const int N = 100010;

typedef long long LL;

LL n, m, res;
LL a[N][6], f[N][6];
unordered_map<int, int> mp;

signed main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
	cin >> n;
	for(int i = 1; i <= n; i ++ ) 
	{
		LL t, x, d;
		cin >> t >> x >> d;
		a[t][x] = d;
	}
	memset(f, -0x3f, sizeof f);
	f[0][0] = 0;
	for(int i = 1; i <= 100000; i ++ ) 
	{
		for(int j = 0; j <= 4; j ++ )
		{
			f[i][j] = a[i][j] + f[i - 1][j];
			if(j > 0) f[i][j] = max(f[i][j], a[i][j] + f[i - 1][j - 1]);
			if(j < 4) f[i][j] = max(f[i][j], a[i][j] + f[i - 1][j + 1]);
		}
	}
	for(int i = 0; i <= 4; i ++ ) 
		res = max(res, f[100000][i]);
	cout << res << endl;
}