第三届河南省acm省赛 BUYING FEED

痛定思痛。 2022-08-21 14:41 125阅读 0赞

##  ##

## 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

2 5 3

3 1 2

4 1 2

1 1 1

样例输出

来源

[第三届河南省程序设计大赛][Link 1]

上传者

[ACM\_赵铭浩][ACM]

题意：

有n个地点有东西

下面是每个地点包含的价格和个数，和哪个地点；

求出到拿m个物品到终点的 最小价格；

\#include <cstdio>  
\#include <cstring>  
\#include <algorithm>  
using namespace std;  
struct node  
\{  
    int cost;  
    int num;  
\}e\[4000\];  
int K, E, N;  
int X\[400\], F\[400\], C\[400\];  
int cmp(node a, node b)  
\{  
    return a.cost < b.cost;  
\}  
int main()  
\{  
    int n;  
    scanf("%d", &n);  
    while(n--)  
    \{  
        scanf("%d %d %d", &K, &E, &N);  
        for(int i = 1; i <= N; i++)  
        \{  
            scanf("%d %d %d", &X\[i\], &F\[i\], &C\[i\]);  
            e\[i\].cost = C\[i\] + E - X\[i\];//计算从这个点买一个到终点所用的价钱。  
            e\[i\].num = F\[i\];//记录这个物品的数量。  
        \}  
        sort(e + 1, e + 1 + N, cmp);  
        /\*for(int i = 1; i <= N; i++)  
        \{  
            printf("%d %d\\n", e\[i\].cost, e\[i\].num);  
        \}\*/  
        long long sum = 0, num = 0;  
        for(int i = 1; i <= N; i++)//贪心  
        \{  
            for(int j = 1; j <= e\[i\].num; j++)  
            \{  
                sum += e\[i\].cost;  
                num++;  
                if(num == K)  
                    break;  
            \}  
            if(num == K)  
                break;  
        \}  
        printf("%lld\\n", sum);  
  
  
    \}  
  
  
    return 0;  
\}

[Link 1]: http://acm.nyist.net/JudgeOnline/search_result.php?source=%E7%AC%AC%E4%B8%89%E5%B1%8A%E6%B2%B3%E5%8D%97%E7%9C%81%E7%A8%8B%E5%BA%8F%E8%AE%BE%E8%AE%A1%E5%A4%A7%E8%B5%9B
[ACM]: http://acm.nyist.net/JudgeOnline/profile.php?userid=ACM_%E8%B5%B5%E9%93%AD%E6%B5%A9

第三届河南省acm省赛 BUYING FEED

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