1033. To Fill or Not to Fill (25)

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.

Input Specification:

Each input file contains one test case. For each case, the first line contains 4 positive numbers: C_max (<= 100), the maximum capacity of the tank; D (<=30000), the distance between Hangzhou and the destination city; D_avg (<=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: P_i, the unit gas price, and D_i(<=D), the distance between this station and Hangzhou, for i=1,...N. All the numbers in a line are separated by a space.

Output Specification:

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.

Sample Input 1:

50 1300 12 86.00 12507.00 6007.00 1507.10 07.20 2007.50 4007.30 10006.85 300

Sample Output 1:

749.17

Sample Input 2:

50 1300 12 27.10 07.00 600

Sample Output 2:

The maximum travel distance = 1200.00

#include <iostream>#include <cstdio>#include <algorithm>#include <limits.h>using namespace std;struct Station{float p;int d;bool operator <(const Station &station) const{return d < station.d;}};Station station[500];int main(void){int cmax, d, davg, n;scanf("%d%d%d%d", &cmax, &d, &davg, &n);for (int i = 0; i < n; ++i)scanf("%f%d", &station[i].p, &station[i].d);sort(station, station + n);if (n <= 0 || station[0].d > 0){printf("The maximum travel distance = 0.00\n");return 0;}int oneD = cmax * davg, nowMax;//加满油后能行驶的最远距离float minP = 0;int first, minIndex;int i = 0, left = 0;while (i < n){/*从第i个汽油站开始，出现的情况是， *1、在行驶最远距离（station[i].d + oneD)没汽油站，则break； *2、在行驶最远距离内有比第i-1站汽油价格更便宜或相等的，找到第一个满足这个条件的，则我们知道了i汽油站需要加多少 *3、在行驶最远距离内都比第i-1站汽油价格高，如果第i-1汽油站加的油能直接到终点，则全部在第i站加油，如不能则选择最便宜的汽油站，在第i-i站加满油 */nowMax = station[i].d + oneD;minIndex = i + 1;first = -1;for (int j = minIndex; j < n && station[j].d <= nowMax && station[j].d < d; ++j){if (station[j].p < station[minIndex].p)minIndex = j; //if (station[j].p <= station[i].p){first = j;break;//}}if (first != -1){//找到了第一个比自己小的minP += (station[first].d - station[i].d - left) * station[i].p;i = first;left = 0;}else{if (minIndex == n){if (nowMax < d)printf("The maximum travel distance = %.2f\n", nowMax * 1.0);else{minP += (d - station[i].d - left) * station[i].p;printf("%.2f\n", minP/davg);}return 0;}else{if (nowMax < d){minP += (oneD - left) * station[i].p;left = nowMax - station[minIndex].d;i = minIndex;}else{minP += (d - station[i].d - left) * station[i].p;printf("%.2f\n", minP / davg);return 0;}}}}return 0;}