DP

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_hour_i ≤ N), an ending hour (starting_hour_i < ending_hour_i ≤ N), and a corresponding efficiency (1 ≤ efficiency_i ≤ 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_hour_i , ending_hour_i , and efficiency_i

Output

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

Sample Input

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

Sample Output

　43

 

---

题意：农夫在M个时间段内可以挤奶（起始时间ST，终止时间ET，该时间段内可以获取的奶量C），奶牛每次产奶后需要休息R小时,求在给定时间，求奶牛在这段时间内的最大产奶量。思路：每次产奶后需要休息R个小时，即挤奶的时间段可视为ST~ET+R。

　　1、将区间按时间段的开始时间进行排序。

　　2、建立数组s[]表示包含本区间并以此区间结尾的区间的最大挤奶量   如 s[3]即为包含有p[3].st~p[3].et区间并以p[3].st~p[3].et结尾的区间的最大挤奶量

　　3、s[]的递推公式：　　若此区间可与前面的区间相接，则保留这个挤奶量，并最终将这些挤奶量的最大值存于数组是s[]

　　　　s[]=max{可与当前区间相接的前面区间的s[]+当前区间的挤奶量};

　　4、遍历数组s[]求数组S中所存的最大值

 

---

 

 

1 #include<cstdio> 2 #include<algorithm> 3 using namespace std; 4 int s[1005]; 5 struct Node{ 6 int st,et,c; 7 }p[1005]; 8 bool cmp( Node a, Node b){ 9 if( a.st<b.st ) 10 return true; 11 return false; 12 } 13 void dp( int m){ 14 for( int i=0; i<m; i++){ 15 s[i]=p[i].c; 16 for( int j=0; j<i; j++){ 17 if( p[j].et<=p[i].st ) 18 s[i]=max(s[i],s[j]+p[i].c); 19 } 20 } 21 } 22 int main() 23 { 24 int n,m,r; 25 int max; 26 27 while(~scanf("%d%d%d",&n,&m,&r)){ 28 max=-1; 29 for( int i=0; i<m; i++){ 30 scanf("%d%d%d",&p[i].st,&p[i].et,&p[i].c); 31 p[i].et+=r; 32 } 33 sort(p,p+m,cmp); 34 dp(m); 35 for( int i=0; i<m; i++) 36 if( s[i]>max ) 37 max=s[i]; 38 printf("%d\n",max); 39 } 40 41 return 0; 42 } View Code

 

 

 

转载于:https://www.cnblogs.com/konoba/p/11287050.html