P2616 [USACO10JAN]購(gòu)買飼料II Buying Feed, II

題目描述

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 FJ 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.

FJ 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 FJ has to pay to buy and transport the K pounds of feed? FJ knows there is a solution.

Consider a sample where FJ 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 FJ 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 FJ travels from 3 to 4 he is moving 1 unit of length and he has 1 pound of feed so he must pay 1*1 = 1 cents.

When FJ travels from 4 to 5 he is 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.

FJ開車去買K份食物，如果他的車上有X份食物。每走一里就花費(fèi)X元。 FJ的城市是一條線，總共E里路，有E+1個(gè)地方，標(biāo)號(hào)0~E。 FJ從0開始走，到E結(jié)束（不能往回走），要買K份食物。 城里有N個(gè)商店，每個(gè)商店的位置是X_i（一個(gè)點(diǎn)上可能有多個(gè)商店），有F_i份食物，每份C_i元。 問(wèn)到達(dá)E并買K份食物的最小花費(fèi)

輸入輸出格式

輸入格式：

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輸出格式：

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輸入輸出樣例

輸入樣例#1：?復(fù)制 2 5 3 3 1 2 4 1 2 1 1 1 輸出樣例#1：?復(fù)制 7

說(shuō)明

?思路：可以轉(zhuǎn)化為01背包求解，也可以排序后貪心。

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; int k,e,n,tot; int f[10010]; int cost[10010]; int x[110],y[110],z[110]; int main(){scanf("%d%d%d",&k,&e,&n);for(int i=1;i<=n;i++) scanf("%d%d%d",&x[i],&y[i],&z[i]);for(int i=1;i<=n;i++)for(int j=1;j<=y[i];j++)cost[++tot]=z[i]+e-x[i];memset(f,0x7f,sizeof(f));f[0]=0;for(int i=1;i<=tot;i++)for(int j=k;j>=1;j--)f[j]=min(f[j],f[j-1]+cost[i]);cout<<f[k]; }

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轉(zhuǎn)載于:https://www.cnblogs.com/cangT-Tlan/p/7887252.html

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