nyoj 248 BUYING FEED 【贪心】



BUYING FEED

时间限制：3000 ms | 内存限制：65535 KB

难度：4

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
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 pay1*1 = 1 cents.

When John travels from 4 to 5 heis 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.

输入The first line of input contains a number c giving the number of cases that follow

There are multi test cases ending with EOF.

Each case starts with a line containing three space-separated integers: K, E, and N

Then N lines follow :every line contains three space-separated integers: Xi Fi Ci

输出For each case,Output A single integer that is the minimum cost for FJ to buy and transport the feed样例输入

1
2 5 3
3 1 2
4 1 2
1 1 1

分析：

农夫要去镇上买饲料，他从0点出发，然后求出他买到所需要的饲料的最小费用（他把一袋饲料从1运到5需要4美元费用）记住要把运费算进去。

代码：

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<stack>
using namespace std;
struct node
{
int get;
int pay;
}t[1010];
bool cmp(node a,node b)
{
return a.pay<b.pay;
}

int main()
{
int T;
int need,total,n,i,num,prize,sum;
scanf("%d",&T);
while(T--)
{
sum=0;
scanf("%d%d%d",&need,&total,&n);
for(i=0;i<n;i++)
{
scanf("%d%d%d",&num,&t[i].get,&prize);
t[i].pay=(total-num)+prize;
}
sort(t,t+n,cmp);
for(i=0;i<n;i++)
{
if(t[i].get>=need)
{
sum=sum+t[i].pay*need;
break;
}
else{
sum=sum+t[i].get*t[i].pay;
need=need-t[i].get;
}
}
printf("%d\n",sum);
}
return 0;
}

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