Western Subregional of NEERC, Minsk, Wednesday, November 4, 2015 Problem C. Cargo Transportation 暴力...

CargoTransportation问题解析

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#数据结构与算法

本文介绍了一个关于货运运输的问题——CargoTransportation，该问题要求在两种不同类型的卡车中选择最优组合，以最低的成本运送指定吨数的货物。通过枚举算法求解最小费用。

Problem C. Cargo Transportation

题目连接：

http://opentrains.snarknews.info/~ejudge/team.cgi?SID=c75360ed7f2c7022&all_runs=1&action=140

Description

The shipping company offers two types of vehicles for bulk cargo transportation. A truck of the first type
can carry Q1 tons of goods in one trip. A single trip costs P1, and the price does not depend on the vehicle
loading level. For a truck of the second type these values equal to Q2 and P2, respectively.
Find the minimum cost to transfer A tons of cargo.

Input

Input contains five positive integers not greater than a thousand: Q1, P1, Q2, P2, A. The numbers are
separated by spaces.

Output

Output one number: the minimum possible price.

Sample Input

3 20 20 100 21

Sample Output

120

Hint

题意

给你两个物品，问你最少花费多少能够装满A。

题解：

2个物品的背包，所以直接xjb枚举就好了

代码

#include <bits/stdc++.h>
#define rep(a,b,c) for(int (a)=(b);(a)<=(c);++(a))
#define drep(a,b,c) for(int (a)=(b);(a)>=(c);--(a))
#define pb push_back
#define mp make_pair
#define sf scanf
#define pf printf
#define two(x) (1<<(x))
#define clr(x,y) memset((x),(y),sizeof((x)))
#define dbg(x) cout << #x << "=" << x << endl;
const int mod = 772002;
int mul(int x,int y){return 1LL*x*y%mod;}
int qpow(int x , int y){int res=1;while(y){if(y&1) res=mul(res,x) ; y>>=1 ; x=mul(x,x);} return res;}
inline int read(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;}
using namespace std;

int main( int argc , char * argv[] ){
    int Q1=read(),P1=read(),Q2=read(),P2=read(),A=read();
    int ans = 2e9;
    for(int i = 0 ; i <= 2000 ; ++ i){
        int t1 = Q1 * i;
        int t2 = (A - t1 + Q2 - 1) / Q2;
        t2 = max( 0 , t2 );
        int cost = i * P1 + t2 * P2;
        ans = min( ans , cost );
    }
    pf("%d\n",ans);
    return 0;
}