998
NYOJ248-BUYING FEED
BUYING FEED時間限製：3000 ms | 內存限製：65535 KB
難度：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.
輸入
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
輸出
For each case,Output A single integer that is the minimum cost for FJ to buy and transport the feed
樣例輸入
1
2 5 3
3 1 2
4 1 2
1 1 1
樣例輸出
7
來源
第三屆河南省程序設計大賽
題意：一條數軸上有n個商店，第i個商店在Xi的位置，最多可以賣Fi磅feed，每磅Ci元。一個人從起點0開始，終點為E，當他到達E點時，至少要買K磅feed，帶著1磅feed每前進一個單位，就要額外花費1元。求最小花費是多少。
思路：貪心算法，將每一個坐標運送一磅feed到終點E的花費算出來，作為商店的一個屬性（姑且叫它“單價”），之後按照這個“單價”從小到大排序，每次取最大值的feed並計算花費，直至K=0，之後輸出結果即為最小值。
注意輸入格式和memset清0（不清0下次排序會出錯）
AC代碼：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct node
{
int num,weight,money,ever;
}per[1500];
int cmp(node x,node y)
{
if(x.ever!=y.ever) return x.ever<y.ever;
}
int main()
{
int i,j,n,K,E,N,sum;
while(scanf("%d",&n)!=EOF)
{
while(n--)
{
memset(per,0,sizeof(per));
scanf("%d %d %d",&K,&E,&N);
for(i=0;i<N;i++)
{
scanf("%d %d %d",&per[i].num,&per[i].weight,&per[i].money);
per[i].ever=per[i].money+(E-per[i].num);
}
sort(per,per+N,cmp);
i=0;sum=0;
while(K!=0)
{
if(K-per[i].weight>=0)
{
K-=per[i].weight;
sum+=per[i].weight*per[i].ever;
}
else
{
sum+=K*per[i].ever;
K=0;
}
i++;
}
printf("%d\n",sum);
}
}
return 0;
}
最後更新：2017-04-03 08:26:19