POJ 1821 - Fence_should sit in front of the plank si and he may…-优快云博客

本文链接：https://blog.youkuaiyun.com/Sureina/article/details/52489458

本文介绍了一种解决最优涂漆任务分配问题的算法。该问题涉及为一系列连续的木板分配多名工人进行涂漆工作，每名工人有其特定的工作区间及报酬，目标是通过动态规划与优先队列实现最大化总收入。

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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

题意：有 n 块木板，m 个员工来刷，每个员工坐在固定的 Si 块木板前面，能刷 Li 长度的木板，得到 Pi 的回报。问能得到的最大回报。

ans[i][j] 表示前 j 块木板截止到第 i 个员工得到的最大回报。

有3种状态转移：1.前 j 块木板由 i-1 个员工完成。

2.前 j-1 块木板由 i 个员工完成。

3.在第 k 块木板的时候由第 i 个员工接手，一直刷到 j 块木板。

然后使用优先队列进行DP优化，每次将在能够刷到范围的状态push进去，每次取最大收益。

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;

struct data
{
    int val, pos;
    bool operator < (const data &d) const
    {
        if (val != d.val)
            return val < d.val;
        else
            return pos > d.pos;
    }
    data() {}
    data(int v, int p)
    {
        val = v;
        pos = p;
    }
};
data tmp[16000 + 5];

struct person
{
    int len, pri, sit;
    bool operator < (const person &p)
    {
        return sit < p.sit;
    }
};
person peo[100 + 5];

int ans[100 + 5][16000 + 5];

int main()
{
    int n, m;

    while (scanf("%d%d", &n, &m) != EOF)
    {
        for (int i = 1; i <= m; ++i)
            scanf("%d%d%d", &peo[i].len, &peo[i].pri, &peo[i].sit);
        peo[m+1].pri = 0;

        sort(peo+1, peo+m+1);
        memset(ans, 0, sizeof(ans));

        for (int i = 1; i <= m; ++i)
        {
            int PRI = peo[i].pri;
            int LEN = peo[i].len;
            int SIT = peo[i].sit;
            priority_queue<data> Q;

            for (int j = max(SIT - LEN, 0); j < SIT; j++)
                Q.push(data(ans[i-1][j] - j*PRI, j));
            for (int j = 1; j <= n; ++j)
            {
                ans[i][j] = max(ans[i-1][j], ans[i][j-1]);

                if (j < SIT || SIT + LEN + 1 < j)
                    continue;

                while (!Q.empty() && Q.top().pos + LEN < j)
                    Q.pop();

                if (Q.empty())
                    continue;
                ans[i][j] = max(ans[i][j], Q.top().val + PRI*j);
            }
        }
        printf("%d\n", ans[m][n]);
    }
    return 0;
}