英文题不出意外还是队友写的， 题意是第一行输入 要买的量k 要走的总长度e 店铺的数量n ； 然后就是每家店铺距离起点的距离 存货量 成本 。
Farmer John needs to travel to town to pick up K (1 <= K <= 100)pounds of feed. Driving D miles with K pounds of feed in his truck costs D*K cents.
The county feed lot has N (1 <= N <= 100) stores (conveniently numbered 1..N) that sell feed. Each store is located on a segment of the X axis whose length is E (1 <= E <= 350). Store i is at
location X_i (0 < X_i < E) on the number line and can sell John as much as F_i (1 <= F_i <= 100) pounds of feed at a cost of C_i (1 <= C_i <= 1,000,000) cents per pound. Amazingly, a given point on the X axis might have more than one store.
Farmer John starts at location 0 on this number line and can drive only in the positive direction, ultimately arriving at location E, with at least K pounds of feed. He can stop at any of the feed stores along the way and buy any amount of feed up to the the store‘s limit.
What is the minimum amount Farmer John has to pay to buy and transport the K pounds of feed? Farmer John knows there is a solution.
Consider a sample where Farmer John needs two pounds of feed from three stores (locations: 1, 3, and 4) on a number line whose range is 0..5:
0 1 2 3 4 5
---------------------------------
1 1 1 Available pounds of feed
1 2 2 Cents per pound
It is best for John to buy one pound of feed from both the second and third stores. He must pay two cents to buy each pound of feed for a total cost of 4. When John travels from 3 to 4 he is moving 1 unit of length and he has 1 pound of feed so he must pay 1*1 = 1 cents.
When John travels from 4 to 5 he is moving one unit and he has 2 pounds of feed so he must pay 1*2 = 2 cents. The total cost is 4+1+2 = 7 cents.
Line 1: Three space-separated integers: K, E, and N
Lines 2…N+1: Line i+1 contains three space-separated integers: Xi Fi Ci
1 #include<stdio.h>
2 #include<stdlib.h>
3 #include<string.h>
4 #include<ctype.h>
5 #define N 1010
6 #define max(a, b)(a > b ? a : b)
7
8 typedef struct node
9 {
10 int d, t, p;
11 int money;
12 } node;
13
14 int cmp(const void *a, const void *b)
15 {
16 node *s1 = (node *)a, *s2 = (node *)b;
17 return s1->money - s2->money;
18 }
19 int main()
20 {
21 node node[N];
22 int k, m, n, sp, st, i;
23 while(scanf("%d%d%d", &k, &m, &n) != EOF)
24 {
25 st = sp = 0;
26 for(i = 0 ; i < n ; i++)
27 {
28 scanf("%d%d%d", &node[i].d, &node[i].t, &node[i].p);
29 node[i].money = (m-node[i].d)+node[i].p;
30 }
31 qsort(node, n, sizeof(node[0]), cmp);
32 for(i = 0; i<n; i++)
33 {
34 if(st < k)
35 {
36 if(node[i].t>k-st || node[i].t == k-st)
37 {
38 sp += (k-st)*node[i].money;
39 break;
40 }
41 else
42 {
43 st += node[i].t;
44 sp += node[i].t*node[i].money;
45 }
46 }
47 }
48 printf("%d\n", sp);
49 }
50 return 0;
51 }
