POJ2431 Expedition

POJ2431 Expedition

Description

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

* 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

* Line 1: 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.

Sample Input

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4 4
5 2
11 5
15 10
25 10

Sample Output

1
2

Hint

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.

Answer

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#include <cstdio>
#include <queue>
#include <algorithm>

typedef std::pair<int, int> S;

const int MAX_N = 10000;

int main() {
int N, L, P;
scanf("%d", &N);
S stops[MAX_N+1];
for (int i = N-1; i >= 0; --i) {
getchar();
scanf("%d %d", &stops[i].first, &stops[i].second);
}
getchar();
scanf("%d %d", &L, &P);

stops[N].first = L;
stops[N].second = 0;
for (int i = 0; i < N; ++i)
stops[i].first = L - stops[i].first;
std::sort(stops, stops+N);

std::priority_queue<int> que;
int d, ans = 0, pos = 0;
for (int i = 0; i <= N; ++i) {
d = stops[i].first - pos;
while (P < d) {
if (que.empty()) {
puts("-1");
return 0;
}
P += que.top();
que.pop();
++ans;
}
P -= d;
pos = stops[i].first;
que.push(stops[i].second);
}
printf("%d\n", ans);
return 0;
}

Analysis

  1. 输入的加油站距离是从城镇(终点)到加油站,并非是卡车所在位置(起点)。
  2. 输入的每个加油站并非是排好序的,需自己按距离排序。
  3. 可以看作,卡车可以在任意时间,加上已经经过的加油站的油。
  4. 每当卡车将车内汽油用完时,选择已经经过的加油站中加油量最大的进行加油。看是否能到达下一个加油站。
  5. 则将已经经过的加油站的油量作为优先队列保存,每次取出最大值。
# POJ

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