經典的貪心演算法 Journey with Pigs (POJ 3544)

Description

Farmer John has a pig farm near town A. He wants to visit his friend living in town B. During this journey he will visit n small villages so he decided to earn some money. He tooks n pigs and plans to sell one pig in each village he visits.

Pork prices in villages are different, in the j-th village the people would buy a pork at pj rubles per kilogram. The distance from town A to the j-th village along the road to town B is dj kilometers.

Pigs have different weights. Transporting one kilogram of pork per one kilometer of the road needs t rubles for addition fuel.

Help John decide, which pig to sell in each town in order to earn as much money as possible.

Input

The first line of the input file contains integer numbers n (1 ≤ n ≤ 1000) and t (1 ≤ t ≤ 109). The second line contains n integer numbers wi (1 ≤ wi ≤ 109)
— the weights of the pigs. The third line contains n integer numbersdj (1 ≤ dj ≤ 109) — the distances to the villages from the town A. The fourth line contains n integer numbers pj (1
≤ pj ≤ 109) — the prices of pork in the villages.

Output

Output n numbers, the j-th number is the number of pig to sell in the j-th village. The pigs are numbered from 1 in the order they are listed in the input file.

Sample Input

3 110 20 1510 20 3050 70 60

Sample Output

3 2 1

#include <iostream>#include <fstream>#include <algorithm>using namespace std;typedef long long int64;struct Pig{int64 value;int pos;};bool cmp(const Pig & p1,const Pig & p2){return p1.value < p2.value;}int64 n,t;/* pigsW  pigsVal 儲存的資料代表這不同的含義 */Pig pigsW[1002]; //每隻pig的 重量 和 標號Pig pigsVal[1002];//每隻pig的 在位置pos處 的  收益int64 d[1002],price;int ans[1002];int main() {int64 i;//ifstream cin("D:\\C++\\algo_poj\\豬_poj_3544\\in.txt");while(cin >> n >> t){for(i=1; i<=n; i++){cin >> pigsW[i].value;pigsW[i].pos = i;}for(i=1; i<=n; i++)cin >> d[i];for(i=1; i<=n; i++){cin >> price;pigsVal[i].value = price - d[i]*t; //每斤豬肉(針對所有豬而言)在位置i處的收益pigsVal[i].pos = i;}sort(pigsW+1, pigsW +1+ n,cmp); //先按照 豬重量排序sort(pigsVal+1, pigsVal +1+ n,cmp); //再按照每個位置的收益排序//重量大的豬 在 收益高的位置for(i=1; i<=n; i++)ans[ pigsVal[i].pos ] = pigsW[i].pos; //可以保證每個位置被用一次。根據貪心策略，可以正確求解for(i=1; i<n; i++)printf("%d ",ans[i]);printf("%d\n",ans[i]);}return 0;}