ZOJ 2109 fatmouse& #39; Trade (Backpack DP + greedy)

Links: 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 he favorite food, JavaBean.
The warehouse has N rooms. The i-th contains j[i] pounds of JavaBeans and requires f[i] pounds of cat food. Fatmouse does not has the to trade for all the JavaBeans in the the the same, instead, he may get j[i]* a% of pounds JavaBeans if he Pays f[i]* a% pounds of cat food. Here A is a real number. Now he's 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 is begins with a line containing the non-negative integers M and N. Then N lines follow, each contains, non-negative integers j[i] and f[i] respectively. The last test case was followed by Two-1 ' s. All integers is 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 J Avabeans 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

Translation:

There used to be only fat rats. His name is Fatmouse, he is like a human terrorist trading arms with the enemy, wretched he prepared a M-pound cat food, ready to trade with the big cats guarding the warehouse, the warehouse has his favorite food JavaBean.

There are n rooms in the warehouse, and in room j[i] pounds javabean and need to f[i] pound cat food exchange, fatmouse do not have to be in every room JavaBean is practical for trading, on the contrary. He can pay the big cat f[i]*a% pounds of cat food, thus changing the j[i]*a% pound of JavaBean. A is a real number, and now he gives you a homework assignment, please tell him how many pounds javabean he can get.
Input descriptive narrative:
The input includes multiple sets of test data, the first line of each test data is two non-negative integer m, N. In the next n rows, each row consists of two nonnegative integers j[i] and f[i], and the last set of test data is two-1. The value of all integers does not paste more than 1000.
Output Descriptive narrative:
For each set of test data, a real number of 3 decimal places is printed on one line, which is the maximum amount of javabean that fatmouse can trade.

Problem Solving Ideas:
The maximum volume of output is required. and retains three decimal places. In this way, we use j[i] divided by f[i] to get a, then, when trading, in order to obtain the most javabean, we must first trade a large. This ensures that the JavaBean can be traded to the maximum.
The data is read into the structure, then the structure as the elements of the vector, and then a from a large to a small order to the vector, and then the calculation in turn. This topic belongs to the Knapsack class topic!

(DP + greedy)

Code:

#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 (M    1.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;}

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ZOJ 2109 fatmouse& #39; Trade (Backpack DP + greedy)