【DP】POJ_3616_Milking Time
来源:互联网 发布:c语言中结构体的使用 编辑:程序博客网 时间:2024/06/03 21:04
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 21 2 810 12 193 6 247 10 31
Sample Output
43
Source
#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn=1e6+10;int dp[maxn];struct node { int l,r,v;}p[maxn];int cmp(struct node a,struct node b){ return a.r<b.r;}int main(){ int n,m,r; scanf("%d%d%d",&n,&m,&r); for(int i=0;i<m;i++) scanf("%d%d%d",&p[i].l,&p[i].r,&p[i].v); sort(p,p+m,cmp); memset(dp,0,sizeof(dp)); for(int i=0;i<m;i++){ dp[i]=p[i].v; if(p[i].r>n) break; for(int j=0;j<m;j++){ if(p[j].r+r<=p[i].l) dp[i]=max(dp[j]+p[i].v,dp[i]); } } int ans=0; for(int i=0;i<m;i++) ans=max(ans,dp[i]); printf("%d\n",ans); return 0;}
- 【DP】POJ_3616_Milking Time
- Milking Time 简单DP
- POJ3616 Milking Time 【DP】
- POJ1316 Milking Time【dp】
- POJ3616 Milking Time DP
- Milking Time(DP)
- poj3616 Miking Time dp
- POJ3616Milking Time(DP)
- POJ3616 Milking Time (dp)
- poj 3616 Milking Time DP
- POJ 3616 Milking Time DP
- Milking Time(3616) 简单dp
- poj-3616 Milking Time 【DP】
- poj 3616 Milking Time DP
- DP-POJ-3616-Milking Time
- poj 3616 Milking Time 【dp】
- [POJ 3616]Milking Time[DP]
- poj3616 Milking Time 入门dp
- hdu 6143 组合数学+dp
- ImmutableMap
- 字符串冒泡排序
- hdu 6143 (2017多校联赛8-11)
- 十个MySQL 数据库经典错误
- 【DP】POJ_3616_Milking Time
- 比特币量化价格浮标-CCI策略
- Html中CSS记录
- CSS布局方式
- Android string.xml 转义字符
- 机房收费系统(一)---登录时出现的问题
- 新闻频道
- 优化Mysql数据库的方法 ? ?
- 字符串面试题(一)字符串逆序