zoukankan      html  css  js  c++  java
  • poj 3616 Milking Time(dp)

    Description

    Bessie is such a hard-working cow. In fact, she is so focused on maximizing her productivity that she decides to schedule her next N (1 ≤ N ≤ 1,000,000) hours (conveniently labeled 0..N-1) so that she produces as much milk as possible.
    
    Farmer John has a list of M (1 ≤ M ≤ 1,000) possibly overlapping intervals in which he is available for milking. Each interval i has a starting hour (0 ≤ starting_houri ≤ N), an ending hour (starting_houri < ending_houri ≤ N), and a corresponding efficiency (1 ≤ efficiencyi ≤ 1,000,000) which indicates how many gallons of milk that he can get out of Bessie in that interval. Farmer John starts and stops milking at the beginning of the starting hour and ending hour, respectively. When being milked, Bessie must be milked through an entire interval.
    
    Even Bessie has her limitations, though. After being milked during any interval, she must rest R (1 ≤ R ≤ N) hours before she can start milking again. Given Farmer Johns list of intervals, determine the maximum amount of milk that Bessie can produce in the N hours.

    Input

    * Line 1: Three space-separated integers: N, M, and R
    * Lines 2..M+1: Line i+1 describes FJ's ith milking interval withthree space-separated integers: starting_houri , ending_houri , and efficiencyi

    Output

    * Line 1: The maximum number of gallons of milk that Bessie can product in the N hours

    Sample Input

    12 4 2
    1 2 8
    10 12 19
    3 6 24
    7 10 31

    Sample Output

    43

    Source

     
     
     
    dp,注意初始化,

    for(int i=1;i<=m;i++){
    dp[i]=cows[i].c;
    }

     1 #include<iostream>
     2 #include<cstdio>
     3 #include<cstring>
     4 #include<queue>
     5 #include<stdlib.h>
     6 using namespace std;
     7 #define N 1000006
     8 #define M 1006
     9 int n,m,r;
    10 struct Node{
    11     int s,e;
    12     int c;
    13 }cows[M];
    14 bool cmp(Node a,Node b){
    15     if(a.s!=b.s)
    16       return a.s<b.s;
    17      return a.e<b.e;
    18 }
    19 int dp[M];//dp[i]表示取到第i段时间段时的最大值 
    20 int main()
    21 {
    22     while(scanf("%d%d%d",&n,&m,&r)==3){
    23         for(int i=1;i<=m;i++){
    24             scanf("%d%d%d",&cows[i].s,&cows[i].e,&cows[i].c);
    25         }
    26         sort(cows+1,cows+m+1,cmp);
    27         memset(dp,0,sizeof(dp));
    28         for(int i=1;i<=m;i++){
    29             dp[i]=cows[i].c;
    30         }
    31         //dp[1]=cows[1].c;
    32         int ans=0;
    33         for(int i=1;i<=m;i++){
    34             for(int j=1;j<i;j++){
    35                 if(cows[i].s>=cows[j].e+r){
    36                     dp[i]=max(dp[i],dp[j]+cows[i].c);
    37                 }
    38             }
    39             ans=max(dp[i],ans);
    40         //    printf("---%d
    ",ans);
    41         }
    42         printf("%d
    ",ans);
    43     }
    44     return 0;
    45 }
    View Code
  • 相关阅读:
    发一弹
    压缩图片
    页面返回并刷新页面
    贤心的WEB弹窗挺不错的
    SMS短信发送API 以后可以弄个短信验证了
    <many-to-one>的fetch属性
    AJAX技术
    中国土地所有权的属性
    Today 's check:mappingResource属性和mappingDirectoryLocations属性的使用
    Pattern类的中文版 菜鸟翻译 有错请纠
  • 原文地址:https://www.cnblogs.com/UniqueColor/p/4800614.html
Copyright © 2011-2022 走看看