第三届河南省程序设计大赛-NYOJ-248-BUYINGFEED

BUYINGFEED
时间限制: 3000ms内存限制: 128000KB 64位整型: Java 类名:
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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 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

典型的贪心问题，不过单价要改成输入的单价加上当前位置到终点的距离，这样对单价排序即可。

代码

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<queue>#include<math.h>#include<algorithm>#include<iostream>using namespace std;//贪心const int maxn=500;struct node{    int Xi;//商店坐标    int Fi;//商店存货量    int Ci;//真实的单价} num[maxn];int K;//最少购买重量为Kint E;//数轴长度int N;//商店数量int cost;//最终花费bool cmp(node x,node y){    return x.Ci<y.Ci;}int main(){    int C;//C组数据    while(~scanf("%d",&C))    {        while(C--)        {            scanf("%d%d%d",&K,&E,&N);            for(int i=1; i<=N; i++)            {                scanf("%d%d%d",&num[i].Xi,&num[i].Fi,&num[i].Ci);                num[i].Ci=num[i].Ci+E-num[i].Xi;//真实单价            }            sort(num+1,num+N,cmp);            cost=0;//初始化总花费            for(int i=1; i<=N&&K>0; i++)            {                cost+=K<=num[i].Fi?K*num[i].Ci:num[i].Fi*num[i].Ci;                K-=num[i].Fi;            }            printf("%d\n",cost);        }    }    return 0;}