[ABC274F] Fishing

Problem Statement

On a number line, there are $N$ fish swimming.

Fish $i$, which has a weight of $W_i$, is at the coordinate $X_i$ at time $0$ and moves at a speed of $V_i$ in the positive direction.

Takahashi will choose an arbitrary real number $t$ greater than or equal to $0$ and do the following action at time $t$ just once.

Action: Choose an arbitrary real number $x$. Catch all fish whose coordinates are between $x$ and $x+A$, inclusive.

Find the maximum total weight of fish that he can catch.

Constraints

• $1 \leq N \leq 2000$
• $1 \leq A \leq 10^4$
• $1 \leq W_i\leq 10^4$
• $0 \leq X_i\leq 10^4$
• $1 \leq V_i\leq 10^4$
• All values in the input are integers.

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Input

The input is given from Standard Input in the following format:

$N$ $A$
$W_1$ $X_1$ $V_1$
$W_2$ $X_2$ $V_2$
$\vdots$
$W_N$ $X_N$ $V_N$

Output

Print the answer.

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Sample Input 1

3 10
100 0 100
1 10 30
10 20 10

Sample Output 1

111

At time $0.25$, fish $1$, $2$, and $3$ are at the coordinates $25$, $17.5$, and $22.5$, respectively. Thus, the action done at this time with $x=16$ catches all the fish.

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Sample Input 2

3 10
100 100 100
1 10 30
10 20 10

Sample Output 2

100

One optimal choice is to do the action at time $0$ with $x=100$.

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Sample Input 3

4 10
1000 100 10
100 99 1
10 0 100
1 1 1

枚举我们选择的第一条鱼。那么如果现在的时间是 \(t\)，如果 \(t\) 秒时一条鱼在这条鱼的右边 \(a\) 格以内，那么这条鱼就能被选中。

称这条鱼的右边 \(a\) 格为答案区间，我们可以算出剩下每条鱼进入答案区间的时间和出来的时间，从而算出每条鱼被算入答案的时间区间。其实就是根据速度大小和差了多远去判断。当然，如果一直不在，就不理这条鱼。如果一直都在，就把这个区间设为 \([1,10000]\)。然后现在要选定一个时间，让他被覆盖的次数最多。那么把所有时间离散化，差分就可以了。

#include<bits/stdc++.h>
using namespace std;
const int N=2005;
int n,x[N],w[N],v[N],c[N<<1],m,ans;
double lsh[N<<1];
struct node{
	double l,r;
	int w;
}xd[N];
int abss(int x)
{
	return x<0? -x:x;
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)
		scanf("%d%d%d",w+i,x+i,v+i);
	for(int i=1;i<=n;i++)
	{
		int k=0,idx=0;
		memset(c,0,sizeof(c));
		for(int j=1;j<=n;j++)
		{
			if(v[i]==v[j])
			{
				if(x[j]>=x[i]&&x[j]-x[i]<=m)
				{
					xd[++k].l=0,xd[k].r=100000;
					lsh[++idx]=0,lsh[++idx]=100000;
					xd[k].w=w[j];
				}
			}
			if(v[i]<v[j]){
				if(x[j]-x[i]>m)
					continue;
				if(x[j]>=x[i])
					xd[++k].l=0;
				else
					xd[++k].l=1.00*(x[i]-x[j])/(v[j]-v[i]);
				xd[k].r=1.00*(x[i]+m-x[j])/(v[j]-v[i]);
				xd[k].w=w[j];
				lsh[++idx]=xd[k].l;
				lsh[++idx]=xd[k].r;
			}
			if(v[i]>v[j])
			{
				if(x[j]<x[i])
					continue;
				if(x[i]+m>x[j])
					xd[++k].l=0;
				else
					xd[++k].l=1.00*(x[j]-x[i]-m)/(v[i]-v[j]);
				xd[k].r=1.00*(x[j]-x[i])/(v[i]-v[j]);
				xd[k].w=w[j];
				lsh[++idx]=xd[k].l;
				lsh[++idx]=xd[k].r;
			}
		}
//		printf("%d\n",k);
		sort(lsh+1,lsh+idx+1);
		for(int j=1;j<=k;j++)
		{
			int l=lower_bound(lsh+1,lsh+idx+1,xd[j].l)-lsh;
			int r=lower_bound(lsh+1,lsh+idx+1,xd[j].r)-lsh;
			c[l]+=xd[j].w,c[r+1]-=xd[j].w;
//			printf("%d %d\n",l,r);
		}
//		putchar('\n');
		for(int j=1;j<=idx;j++)
			c[j]+=c[j-1],ans=max(ans,c[j]);
	}
	printf("%d",ans);
}