poj 1821 dp+单调队列

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#单调队列 #dp

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本文介绍了一种解决最优涂漆区间分配问题的算法，该问题涉及如何为多个工人分配连续的涂漆区间以最大化总收入。文章详细描述了输入输出格式，并提供了一个使用双端队列实现动态规划的具体解决方案。

Fence

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 3858		Accepted: 1176

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

Sample Output

Hint

#include <iostream>
#include <cstring>
#include <cstdio>
#include <deque>
#include <algorithm>

using namespace std;

#define maxn 16001
#define maxk 101
typedef long long ll;

int n, k;
ll dp[maxn];
struct Data
{
    ll l, p, s;
    bool operator <(const Data&a) const
    {
        return s < a.s;
    }
}a[maxk];

int main()
{
    while(~scanf("%d%d", &n, &k)){
        for(int i = 0; i < k; i++)
            scanf("%I64d%I64d%I64d", &a[i].l, &a[i].p, &a[i].s);
        sort(a, a+k);
        memset(dp, 0, sizeof(dp));


        for(int i = 0; i < k; i++){
            deque<int> que;
            for(int j = max(a[i].s-a[i].l, (ll)0); j < a[i].s; j++){
                while(!que.empty()){
                    int last = que.back();
                    if(dp[last]+a[i].p*(i-last) <= dp[j]+a[i].p*(i-j))
                        que.pop_back();
                    else break;
                }
                que.push_back(j);
            }

            for(int j = a[i].s; j < a[i].s+a[i].l && j <= n; j++){
                while(!que.empty() && que.front() < j-a[i].l) que.pop_front();

                if(!que.empty())
                dp[j] = max(dp[j], dp[que.front()]+a[i].p*(j-que.front()));
            }
        }

        ll ans = 0;
        for(int i = 0; i <= n; i++)
            ans = max(ans, dp[i]);
        printf("%I64d\n", ans);
    }
    return 0;
}