题目描述
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 FJ 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.
FJ 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 FJ has to pay to buy and transport the K pounds of feed? FJ knows there is a solution.
Consider a sample where FJ 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 FJ 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 FJ 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 FJ 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.
FJ开车去买K份食物，如果他的车上有X份食物。每走一里就花费X元。 FJ的城市是一条线，总共E里路，有E+1个地方，标号0~E。 FJ从0开始走，到E结束（不能往回走），要买K份食物。 城里有N个商店，每个商店的位置是X_i（一个点上可能有多个商店），有F_i份食物，每份C_i元。 问到达E并买K份食物的最小花费
输入输出格式
输入格式：
输出格式：
输入输出样例
输入样例#1： 复制
2 5 3 3 1 2 4 1 2 1 1 1
输出样例#1： 复制
7
说明
因为题意中所说，车上有X份食物，每走一里就花费X元。也就是说，在整个过程中，因食物份数而产生的花费可以看作每一个食物单独从商店到终点产生的花费
所以可以将一个食物取走的花费+到终点要的花费作为一个成本，然后按花费排序（从小到大）
#include <iostream>
#include<algorithm>
using namespace std;
struct node
{
int x;
int f;
int c;
};
bool com(node fx,node fy)
{
return fx.c<fy.c;
}
int main()
{
node a[150];
int k,e,n;
cin>>k>>e>>n;
for(int i=1;i<=n;i++)
{
cin>>a[i].x>>a[i].f>>a[i].c;
a[i].c+=e-a[i].x;
}
sort(a+1,a+n+1,com);
long long ans=0;
int t=0;
for(int i=1;i<=n;i++)
{
if(t+a[i].f>k)
{
ans+=(k-t)*a[i].c;
break;
}
else
{
ans+=a[i].f*a[i].c;
t+=a[i].f;
}
}
cout<<ans;
return 0;
}