POJ 1821 Fence 单调队列优化DP

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title

POJ 1821
Description

A team of k (1 <= K <= 100) workers should paint a fence which contains N (1 <= N <= 16 000) planks numbered from 1 to N from left to right. Each worker i (1 <= i <= K) should sit in front of the plank Si and he may paint only a compact interval (this means that the planks from the interval should be consecutive). This interval should contain the Si plank. Also a worker should not paint more than Li planks and for each painted plank he should receive Pi $ (1 <= Pi <= 10 000). A plank should be painted by no more than one worker. All the numbers Si should be distinct.
Being the team’s leader you want to determine for each worker the interval that he should paint, knowing that the total income should be maximal. The total income represents the sum of the workers personal income.
Write a program that determines the total maximal income obtained by the K workers.

Input

The input contains:
Input
N K
L1 P1 S1
L2 P2 S2
…
LK PK SK
Semnification
N -the number of the planks; K ? the number of the workers
Li -the maximal number of planks that can be painted by worker i
Pi -the sum received by worker i for a painted plank
Si -the plank in front of which sits the worker i

Output

The output contains a single integer, the total maximal income.

Sample Input

8 4
3 2 2
3 2 3
3 3 5
1 1 7

Sample Output

17

Hint

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Explanation of the sample:
the worker 1 paints the interval [1, 2];
the worker 2 paints the interval [3, 4];
the worker 3 paints the interval [5, 7];
the worker 4 does not paint any plank

Source

Romania OI 2002

analysis

先设出方程，设

$f[i][j]$表示前 $i$个工人，前 $j$个篱笆的最大获利，

那么就有
$f[i][j]=max(f[i-1][j],f[i][j-1]);表示$第 $i$个人不刷，第 $j$面墙不刷

$f[i][j]=max(f[i][j],f[i-1][k]+(j-k)*P[i]);_{ j&gt;=S[i],k&lt;S[i]\&amp;\&amp;k+L[i]&gt;=j}$
那么我们就能想到用单调队列来优化了，

这里主要要想清楚 $k$的范围，那么就好做了。

code

#include<bits/stdc++.h>
using namespace std;
const int maxn=16e3+10,maxk=110;

template<typename T>inline void read(T &x)
{
	x=0;
	T f=1, ch=getchar();
	while (!isdigit(ch) && ch^'-') ch=getchar();
	if (ch=='-') f=-1, ch=getchar();
	while (isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48), ch=getchar();
	x*=f;
}

struct qwq
{
	int l,p,s;
	inline bool operator < (const qwq &a) const
	{
		return s<a.s;
	}
}a[maxk];

int f[maxk][maxn],q[maxn];
inline int calc(int i,int k)
{
	return f[i-1][k]-a[i].p*k;
}

int main()
{
	int n,m;read(n);read(m);
	for (int i=1; i<=m; ++i) read(a[i].l),read(a[i].p),read(a[i].s);

	sort(a+1,a+m+1);
	for (int i=1; i<=m; ++i)
	{
		int l=1,r=0;
		for (int k=max(0,a[i].s-a[i].l); k<=a[i].s-1; ++k)
		{
			while (l<=r && calc(i,q[r])<=calc(i,k)) --r;
			q[++r]=k;
		}

		for (int j=1; j<=n; ++j)
		{
			f[i][j]=max(f[i-1][j],f[i][j-1]);
			if (j>=a[i].s)
			{
				while (l<=r && q[l]<j-a[i].l) ++l;
				if (l<=r) f[i][j]=max(f[i][j],calc(i,q[l])+a[i].p*j);
			}
		}
	}

	printf("%d\n",f[m][n]);
	return 0;
}