BUYING FEED
- 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 poundIt 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
-
题目连接：点击打开链接
题意：有T组测试数据，有个人要去买k重量的东西，但他如果每带k(1....K)重量的东西走L距离时的花费为K*L（不带东西走花费为0），现在要从0点走到E点,有N个点卖东西，问题是要从那个店开始买东西到达E点时的花费最小？
思路：
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。
DP，转移方程（条件不同）
DP[I][J]=MIN{DP[I-1][K]+K*K*(D[I]-D[I-1])+(J-K)*C[I]} J-F[I]<=K<=J
根据题目数据给的范围，500(I的范围)*10000(J的范围)*500(K的范围)肯定要超时。
于是用单调队列优化。
关于单调队列：队头的元素肯定是最小的（最大的）
取元素操作：从队头取出元素，直到该元素在要求的范围内（本题即为J-F[I]<=K<=J）
插入元素操作：若队尾元素比插入的元素要大，则删除。直到比待插入的元素小为止，然后再插入元素。
另外通过本题明确了队列的用法，头指针指向第一个元素，尾指针指向最后一个元素的后一位。若头尾指针相等则队列为空。
代码：#include<iostream>
#include<cstdio>
#include<string>
#include<string.h>
#include<algorithm>
#include<cmath>
#include<cstring>
using namespace std;
struct good
{
int valu;//该点的价值
int newvalu;//新的价值
int nowpiont;//所在的点
int mass;//该点的物品量
}goods[150];
bool cmp(good a,good b)
{
return a.newvalu<b.newvalu;
}
int main()
{
int t;
int x,allpoint ,n;
cin>>t;
while(t--)
{
cin>>x>>allpoint>>n;
for(int i=0;i<n;i++)
{
cin>>goods[i].nowpiont>>goods[i].mass>>goods[i].valu;
goods[i].newvalu=allpoint-goods[i].nowpiont+goods[i].valu;
}
sort(goods,goods+n,cmp);
int sum=0,cnt=0;
for(int i=0;i<n;i++)
{
if(cnt<x)//如果当前的重量不大于总重量
{
if(cnt+goods[i].mass<=x)//当前重量累加小于总重量
{
sum+=goods[i].newvalu*goods[i].mass;//当前的价值乘当前的重量
cnt+=goods[i].mass;//重量累加
if(cnt==x)break;//如过累加后的重量正好等于总重量，break
}
else
{
sum+=(x-cnt)*goods[i].newvalu;//如果当前的重量大于总重量，
break; //只取其中的一部分
}
}
}
cout<<sum<<endl;
}
}