題目地址：http://ac.jobdu.com/problem.php?pid=1437

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題目描述：

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.

輸入：

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.

輸出：

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.

樣例輸入：

50 1300 12 8 6.00 1250 7.00 600 7.00 150 7.10 0 7.20 200 7.50 400 7.30 1000 6.85 300 50 1300 12 2 7.10 0 7.00 600

樣例輸出：

749.17 The maximum travel distance = 1200.00

/** Main.c** Created on: 2014年1月18日* Author: Shaobo* greedy algorithm*/ #include <stdio.h> #include <stdlib.h> #include <math.h>#define MAXN 501 #define MAXC 30000000.0typedef struct station{float price;int dist; }Station;int compare(const void * p, const void * q){Station * p1 = (Station *)p;Station * q1 = (Station *)q;return p1->dist - q1->dist; }int main(void){int Cmax, D, Davg, N; //容量、距離、每單位氣行駛的距離、加氣站總數int i;Station sta[MAXN];float sum, remind_gas, tmp;int k, step;while (scanf("%d %d %d %d", &Cmax, &D, &Davg, &N) != EOF){for (i=0; i<N; ++i){scanf("%f %d", &sta[i].price, &sta[i].dist);}sta[N].dist = D;sta[N].price = 1000000.0;qsort(sta, N, sizeof(Station), compare); //按與杭州距離大小給加氣站排序if (sta[0].dist > 0){printf ("The maximum travel distance = 0.00\n");continue;}sum = 0; //總費用step = Cmax*Davg; //加滿油行駛最大距離remind_gas = 0; //剩余油量for (i=0; i<N; ++i){k = i+1;if (i != 0)remind_gas -= ((float)(sta[i].dist -sta[i-1].dist))/Davg;for (; k<N && sta[k].price>=sta[i].price; ++k)continue;if (sta[k].dist-sta[i].dist > step){sum += (Cmax-remind_gas)*sta[i].price;remind_gas = Cmax;}else{tmp = ((float)(sta[k].dist-sta[i].dist))/Davg - remind_gas;if (fabs(tmp)>1e-5 && tmp>0){sum += tmp*sta[i].price;remind_gas = ((float)(sta[k].dist-sta[i].dist))/Davg;}}if (sta[i+1].dist - sta[i].dist > step){printf ("The maximum travel distance = %.2f\n", (float)(sta[i].dist+step));break;}}if (i == N){printf ("%.2f\n", sum);}}return 0; }

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轉載于:https://www.cnblogs.com/liushaobo/p/4373857.html

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