ZOJ 2109 FatMouse' Trade （背包 dp + 贪婪）

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*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;

}