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  • POJ3616 Milking Time —— DP

    题目链接:http://poj.org/problem?id=3616

    Milking Time
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 10819   Accepted: 4556

    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: NM, 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

     
     
     
    题解:
    经典的类LIS题型(最长上升子序列)。
    类似的题:HDU1160
     
     
    代码如下:
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 #include <cmath>
     5 #include <algorithm>
     6 #include <vector>
     7 #include <queue>
     8 #include <stack>
     9 #include <map>
    10 #include <string>
    11 #include <set>
    12 #define ms(a,b) memset((a),(b),sizeof((a)))
    13 using namespace std;
    14 typedef long long LL;
    15 const double EPS = 1e-8;
    16 const int INF = 2e9;
    17 const LL LNF = 2e18;
    18 const int MAXN = 1e3+10;
    19 
    20 struct node
    21 {
    22     int st, en, val;
    23     bool operator<(const node &x)const{
    24         return st<x.st;
    25     }
    26 }a[MAXN];
    27 int dp[MAXN];
    28 int N, M, R;
    29 
    30 int main()
    31 {
    32     while(scanf("%d%d%d", &N, &M, &R)!=EOF)
    33     {
    34         for(int i = 1; i<=M; i++)
    35             scanf("%d%d%d", &a[i].st, &a[i].en, &a[i].val);
    36 
    37         sort(a+1, a+1+M);
    38         memset(dp, 0, sizeof(dp));
    39         for(int i = 1; i<=M; i++)
    40         for(int j = 0; j<i; j++)
    41             if(j==0 || a[i].st>=a[j].en+R )
    42                 dp[i] = max(dp[i], dp[j] + a[i].val);
    43 
    44         int ans = -INF;
    45         for(int i = 1; i<=M; i++)
    46             ans = max(ans, dp[i]);
    47         printf("%d
    ", ans);
    48     }
    49 }
    View Code
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  • 原文地址:https://www.cnblogs.com/DOLFAMINGO/p/7631265.html
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