POJ 3544 Journey with Pigs：贪心及排序不等式

POJ 3544 Journey with Pigs：贪心及排序不等式2014-07-10Journey with Pigs

http://poj.org/problem？id=3544

Time Limit: 1000MS

Memory Limit: 65536K

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

3 110 20 1510 20 3050 70 60

Sample Output

3 2 1

Source

Northeastern Europe 2007, Northern Subregion

思路：首先根据村庄离A的距离和单位路程的花费，以及当地猪肉的价格，我们可以把到达每一个村庄卖猪单位重量赚的钱算出来。然后，按收益降序排列，再把猪的重量降序排序，这时，根据排序不等式，就可以达到最大盈利了。

维基百科——排序不等式

完整代码：

01./*32ms,216KB*/02.03.#include <cstdio>04.#include <algorithm>05.using namespace std;06.typedef long long ll;07.08.struct Node09.{10.ll value;11.int position;12.bool operator < （const Node a） const13.{14.return value > a.value;15.}16.} weight[1010], earn[1010];17.18.ll dis[1010];19.int ans[1010];20.21.int main（void）22.{23.int n;24.ll t;///单位运费25.while （~scanf（"%d%lld", &n, &t））26.{27.for （int i = 1; i <= n; i++）28.{29.scanf（"%lld", &weight[i].value）;30.weight[i].position = i;31.}32.for （int i = 1; i <= n; i++）33.scanf（"%lld", &dis[i]）;34.for （int i = 1; i <= n; i++）35.{36.ll x;37.scanf（"%lld", &x）;38.earn[i].value = x - dis[i] * t;39.earn[i].position = i;40.}41.sort（weight + 1, weight + 1 + n）;42.sort（earn + 1, earn + 1 + n）;43.for （int i = 1; i <= n; i++）44.ans[earn[i].position] = weight[i].position;45.for （int i = 1; i < n; i++）46.printf（"%d ", ans[i]）;47.printf（"%d
", ans[n]）;48.}49.return 0;50.}