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

题意：John 需要买饲料，有一条X轴，在X轴的某些点上面有店出售饲料。题目给出了有饲料的商店的编号跟每磅饲料需要付出的价钱。各个商店的饲料价钱可能不想同，还要最多能出售的饲料的数目。最主要的是饲料有质量，题目说是从0为起始点，假设在1点买了一磅食物，这一磅食物运到终点假设坐标为5，需要花费（5-1）=4个单位的运费。

分析：求出每个商店（每一点）每买一磅食物需要的价钱，价钱中包括从这点运到终点需要的运费。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

struct P
{
    int pay;      //单位物品支付价钱
    int get;
}p[1010];

int t;
int need, total, n;
int num, prize, sum;

//  按照各点的单位物品支付价钱从小到大排序，为贪心做准备
int cmp(P a, P b)     
{
    return a.pay < b.pay;
}

int main()
{
    scanf("%d", &t);
    while (t--){
        sum = 0;
        scanf("%d%d%d", &need, &total, &n);    //need代表需要的总质量，total代表有在轴上有多少个点
        for (int i = 0; i < n; i++){
            scanf("%d%d%d", &num, &p[i].get, &prize);    //num代表点的编号，prize表示每买单位质量花的价钱
            p[i].pay = (total - num) + prize;    //计算每点每买单位质量需要花的钱
        }
        sort(p, p + n, cmp);      //贪心，每次都取花费最少的
        for (int i = 0; i < n; i++){
            if (p[i].get >= need){
                sum += p[i].pay * need;
                break;
            }
            else{
                sum += p[i].get * p[i].pay;
                need -= p[i].get;
            }
        }
        printf("%d\n", sum);
    }
    return 0;
}