Expedition Problem from SPOJ. Using heap data structure

  algorithm, c++, greedy, heap, max-heap

Problem Expedition

A group of cows grabbed a truck and ventured on an expedition deep
into the jungle. Being rather poor drivers, the cows unfortunately
managed to run over a rock and puncture the truck’s fuel tank. The
truck now leaks one unit of fuel every unit of distance it travels.

To repair the truck, the cows need to drive to the nearest town (no
more than 1,000,000 units distant) down a long, winding road. On this
road, between the town and the current location of the truck, there
are N (1 <= N <= 10,000) fuel stops where the cows can stop to acquire
additional fuel (1..100 units at each stop).

The jungle is a dangerous place for humans and is especially dangerous
for cows. Therefore, the cows want to make the minimum possible number
of stops for fuel on the way to the town. Fortunately, the capacity of
the fuel tank on their truck is so large that there is effectively no
limit to the amount of fuel it can hold. The truck is currently L
units away from the town and has P units of fuel (1 <= P <=
1,000,000).

Determine the minimum number of stops needed to reach the town, or if
the cows cannot reach the town at all.

Input The first line of the input contains an integer t representing the number of test cases. Then t test cases follow. Each
test case has the follwing form:

  • Line 1: A single integer, N
  • Lines 2..N+1: Each line contains two space-separated integers describing a fuel stop: The first integer is the distance from the
    town to the stop; the second is the amount of fuel available at that
    stop.
  • Line N+2: Two space-separated integers, L and P

Output For each test case, output a single integer giving the minimum number of fuel stops necessary to reach the town. If it is not
possible to reach the town, output -1.

Example
Input:

1

4

4 4

5 2

11 5

15 10

25 10

Output:

2

Input details The truck is 25 units away from the town; the truck has
10 units of fuel. Along the road, there are 4 fuel stops at distances
4, 5, 11, and 15 from the town (so these are initially at distances
21, 20, 14, and 10 from the truck). These fuel stops can supply up to
4, 2, 5, and 10 units of fuel, respectively.

Output details: Drive 10 units, stop to acquire 10 more units of fuel,
drive 4 more units, stop to acquire 5 more units of fuel, then drive
to the town.

My attempt

#include <bits/stdc++.h>
using namespace std;
 
int run_case() {
    int N;
    cin >> N;
    vector<pair<int, int>> biasDistance;
    biasDistance.emplace_back(0, 0); // location of tower
    priority_queue<pair<int, int>, vector<pair<int, int>>, less<>> biasFuel;
    for (int i = 0; i < N; ++i) {
        int a, b;
        cin >> a >> b;
        biasDistance.emplace_back(a, b);
    }
    sort(biasDistance.begin(), biasDistance.end());
 
    int L, P;
    cin >> L >> P;
 
    int res = 0;
    while (!biasDistance.empty()) {
        // which points we can reach
        while (P >= L - biasDistance.back().first && !biasDistance.empty()) {
            biasFuel.push({biasDistance.back().second, biasDistance.back().first});
            biasDistance.pop_back();
        }
 
        if (biasDistance.empty()) { // if we can reach all the points
            return res;
        } else { // if not enough fuel to next point --> need stop somewhere
            while (P < L - biasDistance.back().first && !biasFuel.empty()) {
                int addedFuel = biasFuel.top().first;
                int location = biasFuel.top().second;
                biasFuel.pop();
                if (L >= location) {
                    P = P - (L - location) + addedFuel;
                    L = location;
                } else {
                    P += addedFuel;
                }
                res++; // stop here
            }
        }
 
        // still not enough fuel to next point
        if (P < L - biasDistance.back().first) {
            return -1;
        }
    }
 
    return res;
}
 
 
int main() {
    int T;
    cin >> T;
    while (T > 0) {
        cout << run_case() << "n";
        T--;
    }
}

I used a (max heap of fuel) for collecting a list point that the group can reach. And use fuel when the vehicle is out of fuel.

I tried to spend a lot of time to solve this problem. But i failed almost tests with the answer is -1

Don’t know why. Could you help me?

Source: Windows Questions C++

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