HDU1009 FatMouse' Trade 【貪心】

標籤：hdu1009

FatMouse‘ Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 42786    Accepted Submission(s): 14274

Problem DescriptionFatMouse 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.
 

 

InputThe 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.
 

 

OutputFor 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 37 24 35 220 325 1824 1515 10-1 -1

 

 

Sample Output

13.33331.500

題意：給定一個容量為m的背包以及n個物品和每個物品的重量及價值，單個物品可以任意切分，求背包能獲得的最大裝載價值。

題解：可以求出每個物品的單位重量價值，排序，每次選擇單位價值最大的裝包即可。

 

#include <stdio.h>#include <algorithm>#define maxn 1002using std::sort;struct Node{int v, c;double val;} arr[maxn];bool cmp(Node a, Node b){return a.val > b.val;}int main(){//freopen("in.txt", "r", stdin);//freopen("out.txt", "w", stdout);int m, n, i, a, b;double ans;while(scanf("%d%d", &m, &n), m >= 0 && n >= 0){for(i = 0; i < n; ++i){scanf("%d%d", &a, &b);arr[i].val = a * 1.0 / b;arr[i].v = a; arr[i].c = b;}ans = 0;sort(arr, arr + n, cmp);for(i = 0; i < n; ++i){if(m >= arr[i].c){m -= arr[i].c; ans += arr[i].v;}else{ans += m * arr[i].val; break;}}printf("%.3lf\n", ans);}return 0;}