经典的贪心算法 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 p_j rubles per kilogram. The distance from town A to the j-th village along the road to town B is d_j 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 w_i (1 ≤ w_i ≤ 10⁹) — the weights of the pigs. The third line contains n integer numbersd_j (1 ≤ d_j ≤ 10⁹) — the distances to the villages from the town A. The fourth line contains n integer numbers p_j (1 ≤ p_j ≤ 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 1
10 20 15
10 20 30
50 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 存储的数据代表这不同的含义 */
Pig pigsW[1002]; //每只pig的 重量 和 标号
Pig pigsVal[1002];//每只pig的 在位置pos处 的  收益
int64 d[1002],price;
int ans[1002];
int main() {
	int64 i;
	//ifstream cin("D:\\C++\\algo_poj\\猪_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; //每斤猪肉(针对所有猪而言)在位置i处的收益
		pigsVal[i].pos = i;
	}
	sort(pigsW+1, pigsW +1+ n,cmp); //先按照 猪重量排序
	sort(pigsVal+1, pigsVal +1+ n,cmp); //再按照每个位置的收益排序

	//重量大的猪 在 收益高的位置
	for(i=1; i<=n; i++)
		ans[ pigsVal[i].pos ] = pigsW[i].pos; //可以保证每个位置被用一次。根据贪心策略，可以正确求解
	for(i=1; i<n; i++)
		printf("%d ",ans[i]);
	printf("%d\n",ans[i]);

	}
	return 0;
}