To Fill or Not to Fill

                 题目1437：To Fill or Not to Fill

时间限制：1 秒

内存限制：128 兆

特殊判题：否

提交：3577

解决：814

题目描述：

With highways available, driving a car from Hangzhou to any other city is easy. But since the tank capacity of a car is limited, we have to find gas stations on the way from time to time. Different gas station may give different price. You are asked to carefully design the cheapest route to go.

输入：

For each case, the first line contains 4 positive numbers: Cmax (<= 100), the maximum capacity of the tank; D (<=30000), the distance between Hangzhou and the destination city; Davg (<=20), the average distance per unit gas that the car can run; and N (<= 500), the total number of gas stations. Then N lines follow, each contains a pair of non-negative numbers: Pi, the unit gas price, and Di (<=D), the distance between this station and Hangzhou, for i=1,...N. All the numbers in a line are separated by a space.

输出：

For each test case, print the cheapest price in a line, accurate up to 2 decimal places. It is assumed that the tank is empty at the beginning. If it is impossible to reach the destination, print "The maximum travel distance = X" where X is the maximum possible distance the car can run, accurate up to 2 decimal places.

样例输入：

50 1300 12 86.00 12507.00 6007.00 1507.10 07.20 2007.50 4007.30 10006.85 30050 1300 12 27.10 07.00 600

样例输出：

749.17The maximum travel distance = 1200.00

来源：

2012年浙江大学计算机及软件工程研究生机试真题

贪心算法。调BUG俩小时发现是精度问题！以后带小数点的测试数据，老老实实全用浮点型吧。。。。

import java.util.*;public class ToFillOrNotToFill {static private Scanner input;static int Cmax, D, Davg, N;/* * 九度oj1437 Cmax (<= 100), the maximum capacity of the tank; D (<=30000), * the distance between Hangzhou and the destination city; Davg (<=20), the * average distance per unit gas that the car can run; and N (<= 500), the * total number of gas stations. */private static final int INF = 1 << 30;static int next; // 下一个要走的“贪心”位置static Station[] s;public static void main(String[] args) {input = new Scanner(System.in);s = new Station[502];for (int i = 0; i < 502; i++)s[i] = new Station(0, 0);while (input.hasNext()) {Cmax = input.nextInt();D = input.nextInt();Davg = input.nextInt();N = input.nextInt();for (int i = 0; i < N; i++) {float price = input.nextFloat();float distance = input.nextFloat();s[i].price = price;s[i].distance = distance;}Arrays.sort(s, 0, N); // 按照距离从小到大排序/* for( int i = 0 ; i < N ; i++ )  System.out.println(s[i].price+"  "+ s[i].distance);*/ int maxDis = Cmax * Davg; // 最大可走距离float totalCost = 0; // 花钱总数float dis = 0; // 总共已行走的距离float leftO = 0; // 剩余的油int i = 0;if (s[0].distance != 0 || N == 0) {System.out.println("The maximum travel distance = 0.00");continue;}while (true) {//System.out.println("当前的站点："+i+"当前使用金钱："+totalCost+"当前剩余油量："+leftO);float minPrice = findMin(i, maxDis);if (minPrice == INF) {//当前站是最后一站if (s[i].distance + maxDis >= D) {totalCost += ((D - s[i].distance) / Davg - leftO)*1.0 * s[i].price;System.out.printf("%.2f\n", totalCost);break;} else {dis = s[i].distance + maxDis;System.out.printf("The maximum travel distance = %.2f\n", dis);break;}} else if (minPrice >= s[i].price) { // 当前站是可达范围内最便宜的站if (s[i].distance + maxDis >= D) {//当前位置可以直达终点站float needO = (float) ((D - s[i].distance)*1.0 / Davg);if (needO <= leftO) {System.out.printf("%.2f\n", totalCost);} else {totalCost += (needO - leftO)* 1.0 * s[i].price;System.out.printf("%.2f\n", totalCost);}break;} else {//当前位置不能到达终点站，在最便宜的站加满油totalCost += (Cmax - leftO) * s[i].price;leftO = (float) (Cmax - ((s[next].distance - s[i].distance)*1.0 / Davg));i = next;}} else { // 可达范围内有更便宜的站totalCost += ((s[next].distance - s[i].distance)*1.0 / Davg - leftO) * s[i].price;/*System.out.println("s[i].distance:"+s[i].distance);System.out.println("s[next].distance:"+s[next].distance);System.out.println(totalCost+"---"+s[next].distance+"$:"+s[i].price);*/leftO = 0;i = next;}//System.out.println(i);}}}private static float findMin(int i, double range) {// TODO Auto-generated method stubfloat maxDis = (float) (s[i].distance + range);float minPrice = INF;for (int j = i + 1; j < N; j++) {if (s[j].distance <= maxDis && s[j].distance < D) { // 既在可达范围内，又在终点前if (s[j].price <= minPrice) { // 更贵则选择最便宜的minPrice = s[j].price;next = j;}if (minPrice < s[i].price) // 若更便宜则选择最近的一个break;} else { // 超出范围break;}}return minPrice;}}class Station implements Comparable<Station> {float price;float distance;public Station(float price, float distance) {this.distance = distance;this.price = price;}@Overridepublic int compareTo(Station o) {// TODO Auto-generated method stubif (this.distance > o.distance) {return 1;} else if (this.distance == o.distance) {return 0;} else {return -1;}}}

来源：http://ac.jobdu.com/problem.php?pid=1437