Fence
Time Limit: 1000MS   Memory Limit: 30000K
Total Submissions: 4705   Accepted: 1489

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

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 
思路:dp+单调队列;
首先我们要对原来的点按顺序排,然后dp[i][j]表示前i个人喷漆到j个位置结束的最大值,那么转移方程是dp[i][j] = max(dp[i-1][j],dp[i-1][j-s]+s*ans.p);这样n^3肯定不行,然后方程可写为dp[i-1][k]+(j-k)*ans.p=dp[i-1][k]-k*ans.p+j*ans.p,因为第二层循环中的j是不变的,(max(0,j-ans.l)<=k<ans.s),那么ans.l定,ans.s定,当j增大时区间范围减小,然后单调队列维护下最大值即可。复杂度O(n*m);
 1 #include<stdio.h>
2 #include<algorithm>
3 #include<iostream>
4 #include<string.h>
5 #include<stdlib.h>
6 #include<queue>
7 #include<stack>
8 using namespace std;
9 typedef long long LL;
10 typedef struct node
11 {
12 int cost;
13 int id;
14 bool operator<(const node &cx)const
15 {
16 if(cx.cost == cost)return cx.id < id;
17 else return cx.cost>cost;
18 }
19 } ak;
20 typedef struct pp
21 {
22 int l,p,s;
23 } ss;
24 bool cmp(pp p,pp q)
25 {
26 return p.s<q.s;
27 }
28 priority_queue<ak>que;
29 ss ans[105];
30 int dp[105][16005];
31 ak quq[2*16005];
32 int main(void)
33 {
34 int n,m;
35 while(scanf("%d %d",&n,&m)!=EOF)
36 {
37 int j;
38 int i;
39 int maxx = 0;
40 for(i = 1; i <= m; i++)
41 scanf("%d %d %d",&ans[i].l,&ans[i].p,&ans[i].s);
42 sort(ans+1,ans+1+m,cmp);
43 memset(dp,0,sizeof(dp));
44 for(i = 1; i <= m; i++)
45 {
46 int head = 16001;
47 int rail = 16000;
48 for(j = 0; j < ans[i].s; j++)
49 {
50 dp[i][j] = dp[i-1][j];
51 ak acc;
52 acc.cost = dp[i-1][j]-j*ans[i].p;
53 acc.id = j;
54 if(head>rail)
55 quq[--head] = acc;
56 else
57 {
58 ak cpp = quq[rail];
59 while(cpp.cost < acc.cost)
60 {
61 rail--;
62 if(rail<head)
63 {
64 break;
65 }
66 cpp = quq[rail];
67 }
68 quq[++rail] = acc;
69 }
70 maxx = max(maxx,dp[i][j]);
71 }
72 for(j = ans[i].s; j <= min(n,ans[i].l+ans[i].s-1); j++)
73 {
74 dp[i][j] = max(dp[i-1][j],dp[i][j]);
75 int minn = max(0,j-ans[i].l);
76 while(head<=rail)
77 {
78 ak acc = quq[head];
79 if(acc.id < minn)
80 {
81 head++;
82 }
83 else
84 {
85 dp[i][j] = max(dp[i][j],acc.cost+j*ans[i].p);
86 break;
87 }
88 }
89 maxx = max(maxx,dp[i][j]);
90 }
91 for(j = min(n,ans[i].l+ans[i].s-1)+1; j <= n; j++)
92 {
93 dp[i][j] = dp[i-1][j];
94 maxx = max(maxx,dp[i][j]);
95 }}
96 printf("%d\n",maxx);
97 }
98 return 0;}

Fence(poj1821)的更多相关文章

  1. DP重开

    颓了差不多一周后,决定重开DP 这一周,怎么说,学了学trie树,学了学二叉堆,又学了学树状数组,差不多就这样,然后和cdc一番交流后发现,学这么多有用吗?noip的范围不就是提高篇向外扩展一下,现在 ...

  2. 【学习笔记】动态规划—各种 DP 优化

    [学习笔记]动态规划-各种 DP 优化 [大前言] 个人认为贪心,\(dp\) 是最难的,每次遇到题完全不知道该怎么办,看了题解后又瞬间恍然大悟(TAT).这篇文章也是花了我差不多一个月时间才全部完成 ...

  3. [POJ1821]Fence(单调队列优化dp)

    [poj1821]Fence 有 N 块木板从左至右排成一行,有 M 个工匠对这些木板进行粉刷,每块木板至多被粉刷一次.第 i 个工匠要么不粉刷,要么粉刷包含木板 Si 的,长度不超过Li 的连续一段 ...

  4. POJ1821 Fence

    题意 Language:Default Fence Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 6478 Accepted: ...

  5. poj1821 Fence【队列优化线性DP】

    Fence Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 6122   Accepted: 1972 Description ...

  6. POJ1821 Fence 题解报告

    传送门 1 题目描述 A team of $k (1 <= K <= 100) $workers should paint a fence which contains \(N (1 &l ...

  7. poj1821 Fence(单调队列优化dp)

    地址 一排N个木板,M个工匠站在不同位置$S_i$,每个人可以粉刷覆盖他位置的.最长长度为$L_i$木板段,每刷一个有$P_i$报酬.同一木板只刷一次.求最大报酬. 根据每个人的位置dp,设$f[i] ...

  8. $Poj1821\ Fence\ $单调队列优化$DP$

    Poj   Acwing Description 有N块木板等待被M个工匠粉刷,每块木板至多被刷一次.第i个工匠要么不粉刷,要么粉刷包含木块Si的,长度不超过Li的连续的一段木板,每粉刷一块可以得到P ...

  9. poj1821 Fence(dp,单调队列优化)

    题意: 由k(1 <= K <= 100)个工人组成的团队应油漆围墙,其中包含N(1 <= N <= 16 000)个从左到右从1到N编号的木板.每个工人i(1 <= i ...

随机推荐

  1. 重学Git(一)

    一.最最最基础操作 # 初始化仓库 git init # 添加文件到暂存区 git add readme.md # 提交 git commit -m 'wrote a readme file' 二.简 ...

  2. CPU如何同时运行多个进程?

    1 # -*- coding: utf-8 -*- 2 import re 3 mem = [x for x in re.split('[\r|\n]', ''' 4 store a 1 5 add ...

  3. day 03Linux修改命令提示符

    day 03Linux修改命令提示符 昨日回顾 1.选择客户机操作系统: Microsoft Windows # 一次只能安装一台电脑 Linux(推荐) VMware ESX # 服务器版本VNwa ...

  4. Android权限级别(protectionLevel)

    通常情况下,对于需要付费的操作以及可能涉及到用户隐私的操作,我们都会格外敏感. 出于上述考虑以及更多的安全考虑,Android中对一些访问进行了限制,如网络访问(需付费)以及获取联系人(涉及隐私)等. ...

  5. Linux学习 - 数值运算

    1 declare 声明变量类型 declare [+/-] [选项] 变量名 - 给变量设定类型属性 + 取消变量的类型属性 -i 将变量声明为整数型 -x 将变量声明为环境变量(同export) ...

  6. ORACLE DBMS_ROWID包详解

    这个包在11gR2中有11个函数或存储: 1. 根据给定参数返回一个rowid --根据给定参数返回一个rowid FUNCTION rowid_create(rowid_type IN NUMBER ...

  7. ORACLE 获取执行计划的方法

    一.获取执行计划的6种方法(详细步骤已经在每个例子的开头注释部分说明了): 1. explain plan for获取: 2. set autotrace on : 3. statistics_lev ...

  8. shell脚本下载网页图片

    和大家分享一个shell脚本写的图片抓取器.使用方法:img_downloader.sh.使用时在shell下输入:./img_downloader.sh www.baidu.com -d image ...

  9. js 长按鼠标左键实现溢出内容左右滚动滚动

    var nextPress, prevPress; // 鼠标按下执行定时器,每0.1秒向左移一个li内容的宽度 function nextDown() { nextPress = setInterv ...

  10. 【MySQL】学生成绩

    统计每个人的总成绩排名 select stu.`name`,sum(stu.score) as totalscore from stu GROUP BY `name` order by totalsc ...