nyoj 248 BUYING FEED

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

样例输出

7

贪心：

 
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAX 100+10
using namespace std;
int need,len,n;//需要食物,线路总长度,总商店数 
struct record
{
    int x;//坐标
    int sell;//可供购买量
    int cost;//单价 
}num[MAX];
bool cmp(record a,record b)
{
    return len-a.x+a.cost<len-b.x+b.cost;
}
int main()
{
    int t,i,j;
    int sum,have;//总花费 和 已有食物数量 
    int d,carry;//记录选择购买食物的商店之间的距离 
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d",&need,&len,&n);
        for(i=0;i<n;i++)
        {
            scanf("%d%d%d",&num[i].x,&num[i].sell,&num[i].cost);
        }
        sort(num,num+n,cmp);
        sum=0; have=0; 
        for(i=0;i<n;i++)
        {
            if(have==need)
            break;
            if(have+num[i].sell<=need)
            {
                sum+=num[i].sell*num[i].cost;//买食物花费 
                sum+=(len-num[i].x)*num[i].sell;//运食物花费 
                have+=num[i].sell;
            }
            else 
            {
                sum+=(need-have)*num[i].cost;//买食物花费 
                sum+=(len-num[i].x)*(need-have);//运食物花费 
                have=need; 
            }
        }
        printf("%d\n",sum);
    }
    return 0;
}