c++ - 来自 SPOJ 的远征问题。使用堆数据结构

Question

问题 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.

我的尝试

#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--;
    }
}

我使用了(最大燃料堆)来收集小组可以到达的列表点。并在车辆没油时使用燃料。

我试图花很多时间来解决这个问题。但我几乎没有通过测试，答案是 -1

不知道为什么。你能帮帮我吗？

Answer 1

由于我们的容量是无限的，所以我们可以在任何阶段积累多少。我们所做的每一次添加都会扩展我们可以达到的范围。因此，考虑迭代的一种方式可能是:

while range r is too small:
  heap pop the highest fuel stop seen, f
  new_range = r + f
  heap push all stops between r and new_range
  r = new_range