给定一个整数K，该整数K表示在长度为N米的直线路径上行驶的汽车的燃料容量为1升/公吨，并具有两个数组a []和b []，每个数组的大小为M，其中a [i]表示第i个加油站的位置， b [i]表示在该加油站1升燃料的成本。任务是找到从0开始到行尾所需的最低成本。如果无法到达终点，则打印-1 。
例子：
Input: N = 10, K = 10, M = 2, a[] = {0, 1}, b[] = {5, 2}
Output: 23
Explanation:
At the 0th location, fill the car tank with 1 liter of fuel at cost 5. Then, at 1st location, fill 9 liters of fuel at cost 18.
Therefore, the minimum cost required is 23.
Input: N = 10, K = 5, M = 3, a[] = {0, 3, 5}, b[] = {5, 9, 3}
Output: 40
方法：想法是使用Priority Queue和HashMap来存储加油站，以便以最低的成本获得加油站。请按照以下步骤解决问题：
- 初始化一个HashMap(例如地图)，以存储车站的索引及其各自的燃油消耗率。
- 初始化一个优先队列，例如pq，以存储加油站的索引，燃料成本以及所加燃料的升数。
- 初始化两个变量，例如cost，以存储所需的最低成本，并设置flag = false来检查是否有加油站。
- 迭代范围[1，N] ：
- 检查是否有电台。如果发现是真的，则将其插入pq 。
- 移开所有无法加油的站。
- 如果pq为空，则不可能到达行尾。因此，将flag设置为true 。
- 存储成本最低的站点，并更新成本和从该特定站点泵出的升数。
- 再次将其插入pq 。
- 如果标志为true，则打印“ -1”。这意味着没有加油站。因此，不可能到达行尾。
- 否则，打印最低成本以达到该行的末尾。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// For priority_queue
struct Compare {
bool operator()(array a, array b)
{
return a[1] > b[1];
}
};
// Function to calculate the minimum cost
// required to reach the end of Line
void minCost(int N, int K, int M, int a[], int b[])
{
// Checks if possible to
// reach end or not
bool flag = true;
// Stores the stations and
// respective rate of fuel
unordered_map map;
for (int i = 0; i < M; i++) {
map[a[i]] = b[i];
if (i == M - 1 && K < N - a[i]) {
flag = false;
break;
}
else if (i < M - 1 && K < a[i + 1] - a[i]) {
flag = false;
break;
}
}
if (!flag) {
cout << -1;
return;
}
// Stores the station index and cost of fuel and
// litres of petrol which is being fueled
priority_queue, vector >,
Compare>
pq;
int cost = 0;
flag = false;
// Iterate through the entire line
for (int i = 0; i < N; i++) {
// Check if there is a station at current index
if (map.find(i) != map.end()) {
array arr = { i, map[i], 0 };
pq.push(arr);
}
// Remove all the stations where
// fuel cannot be pumped
while (pq.size() > 0 && pq.top()[2] == K)
pq.pop();
// If there is no station left to fill fuel
// in tank, it is not possible to reach end
if (pq.size() == 0) {
flag = true;
break;
}
// Stores the best station
// visited so far
array best_bunk = pq.top();
pq.pop();
// Pump fuel from the best station
cost += best_bunk[1];
// Update the count of litres
// taken from that station
best_bunk[2]++;
// Update the bunk in queue
pq.push(best_bunk);
}
if (flag) {
cout << -1 << "\n";
return;
}
// Print the cost
cout << cost << "\n";
}
// Driven Program
int main()
{
// Given value of N, K & M
int N = 10, K = 3, M = 4;
// Given arrays
int a[] = { 0, 1, 4, 6 };
int b[] = { 5, 2, 2, 4 };
// Function call to calculate minimum
// cost to reach end of the line
minCost(N, K, M, a, b);
return 0;
}
// This code is contributed by Kingash. Java
// Java program for the above approach
import java.util.*;
public class Main {
// Function to calculate the minimum cost
// required to reach the end of Line
static void minCost(int N, int K, int M,
int[] a, int[] b)
{
// Checks if possible to
// reach end or not
boolean flag = true;
// Stores the stations and
// respective rate of fuel
HashMap map = new HashMap<>();
for (int i = 0; i < M; i++) {
map.put(a[i], b[i]);
if (i == M - 1 && K < N - a[i]) {
flag = false;
break;
}
else if (i < M - 1 && K < a[i + 1] - a[i]) {
flag = false;
break;
}
}
if (!flag) {
System.out.println("-1");
return;
}
// Stores the station index and cost of fuel and
// litres of petrol which is being fueled
PriorityQueue pq
= new PriorityQueue<>((c, d) -> c[1] - d[1]);
int cost = 0;
flag = false;
// Iterate through the entire line
for (int i = 0; i < N; i++) {
// Check if there is a station at current index
if (map.containsKey(i))
pq.add(new int[] { i, map.get(i), 0 });
// Remove all the stations where
// fuel cannot be pumped
while (pq.size() > 0 && pq.peek()[2] == K)
pq.poll();
// If there is no station left to fill fuel
// in tank, it is not possible to reach end
if (pq.size() == 0) {
flag = true;
break;
}
// Stores the best station
// visited so far
int best_bunk[] = pq.poll();
// Pump fuel from the best station
cost += best_bunk[1];
// Update the count of litres
// taken from that station
best_bunk[2]++;
// Update the bunk in queue
pq.add(best_bunk);
}
if (flag) {
System.out.println("-1");
return;
}
// Print the cost
System.out.println(cost);
}
// Driver Code
public static void main(String args[])
{
// Given value of N, K & M
int N = 10, K = 3, M = 4;
// Given arrays
int a[] = { 0, 1, 4, 6 };
int b[] = { 5, 2, 2, 4 };
// Function call to calculate minimum
// cost to reach end of the line
minCost(N, K, M, a, b);
}
} 输出：
-1时间复杂度： O(N * logN)
辅助空间： O(N)