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ZOJ 2109 FatMouse' Trade (背包 dp + 贪婪)

时间:2015-07-21 20:32:47      阅读:134      评论:0      收藏:0      [点我收藏+]

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链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1109

FatMouse prepared M pounds of cat food, ready to trade with the cats guarding the warehouse containing his favorite food, JavaBean.
The warehouse has N rooms. The i-th room contains J[i] pounds of JavaBeans and requires F[i] pounds of cat food. FatMouse does not have to trade for all the JavaBeans in the room, instead, he may get J[i]* a% pounds of JavaBeans if he pays F[i]* a% pounds of cat food. Here a is a real number. Now he is assigning this homework to you: tell him the maximum amount of JavaBeans he can obtain.

Input
The input consists of multiple test cases. Each test case begins with a line containing two non-negative integers M and N. Then N lines follow, each contains two non-negative integers J[i] and F[i] respectively. The last test case is followed by two -1‘s. All integers are not greater than 1000.

Output
For each test case, print in a single line a real number accurate up to 3 decimal places, which is the maximum amount of JavaBeans that FatMouse can obtain.

Sample Input
5 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1

Sample Output
13.333
31.500


翻译:

从前有仅仅肥肥的老鼠。他叫FatMouse,他就像人类的恐怖分子跟敌人交易军火一样,猥琐的他准备了M磅猫食,准备与守卫仓库的大猫们进行交易,仓库里有他最爱吃的食物Javabean。


仓库里有N个房间,第i间房间里有J[i]磅Javabean且须要F[i]磅猫食进行交换,FatMouse不必吧每一个房间里的Javabean所实用于交易,相反。他可以付给大猫F[i]*a%磅猫食,从而换的J[i]*a%磅的Javabean。当中,a是一个实数,如今他给你布置一个家庭作业,请你告诉他他最多可以获得多少磅Javabean。
输入描写叙述:
输入包括多组測试数据,每组測试数据的开头一行是两个非负整数M, N.接下来的N行中,每行包括两个非负整数J[i]和F[i],最后一组測试数据是两个-1。全部的整数的值不糊超过1000。
输出描写叙述:
对于每组測试数据,在一行上打印出一个3位小数的实数,这个实数是FatMouse可以交易到的最大数量的Javabean.


解题思路:
本题要求输出最大交易量。并保留三位小数。这样,我们使用J[i]除以F[i]就得到了a,那么,交易的时候,为了获得最多的Javabean,要先交易a大的。这样就确保了能交易到最多的Javabean.
把数据读入结构体中,再将结构体作为向量的元素,再按a由大到小的顺序给向量排序,然后依次进行计算。这样的题目属于背包类的题目!

(dp + 贪心)


代码:

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <set>
#define MAXN 10005
#define RST(N)memset(N, 0, sizeof(N))
#include <algorithm>
using namespace std;

typedef struct Mouse_ {
    double J, F;
    double a;
}Mouse;

int n, m;
vector <Mouse> v;
vector <Mouse> ::iterator it;

bool cmp(const Mouse m1, const Mouse m2)
{
    if(m1.a != m2.a) return m1.a > m2.a;
    else return m1.F < m2.F;
}

int main()
{
    while(~scanf("%d %d", &n, &m)) {
        if(n == -1 && m == -1) break;
        Mouse mouse;
        v.clear();
        for(int i=0; i<m; i++) {
            scanf("%lf %lf", &mouse.J, &mouse.F);
            mouse.a = mouse.J/mouse.F;
            v.push_back(mouse);
        }
        sort(v.begin(), v.end(), cmp);
        double sum = 0;
        for(int i=0; i<v.size(); i++) {
            if(n > v[i].F) {
                sum += v[i].J;
                n -= v[i].F;
            }else {
                sum += n*v[i].a;
                break;
            }
        }
        printf("%.3lf\n", sum);
    }
    return 0;
}


版权声明:本文博客原创文章,博客,未经同意,不得转载。

ZOJ 2109 FatMouse&#39; Trade (背包 dp + 贪婪)

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原文地址:http://www.cnblogs.com/mengfanrong/p/4665440.html

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