POJ - 2431 Expedition（优先队列）

Expedition

Time Limit: 1000MS Memory Limit: 65536KB 64bit IO Format: %I64d & %I64u

Submit Status

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

44 45 211 515 1025 10

Sample Output

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.

题意：一辆卡车距离城镇L单位长度，初始有P油，每行驶一个单位长度消耗一单位油。有n个加油站可以加油，给出n个加油站与城镇的距离Ai，和在加油站可以加的有的量Bi，问最少加油几次才能行驶L长度，如果不能输出-1。

我们稍微变换一下思路：每次经过加油站，都把油装到瓶子里面带走，Bi加入优先队列，到需要加油的时候才加，因为需要使加油的次数最少，每次都是选油量最大的加。如果优先队列为空而没有到达下一加油站或是终点。则不能够到达。

#include<iostream>#include<cstring>#include<cstdio>#include<queue>#include<algorithm>using namespace std;const int MAXN = 10100;int n, l, p;struct tank{int a, b;bool operator < (const tank& bb) const{return a > bb.a;}}t[MAXN];int main(){scanf("%d", &n);for (int i = 0; i < n; i++)scanf("%d%d", &t[i].a, &t[i].b);scanf("%d%d", &l, &p);sort(t, t + n);priority_queue<int>que;int ans = 0;int npos = l;int P = p;t[n].a = 0;t[n].b = 0;for (int i = 0; i <= n; i++){int dis = npos - t[i].a;P = P - dis;while (P < 0){if (que.empty()){cout << -1 << endl;return 0;}P = P + que.top();que.pop();ans++;}npos = t[i].a;que.push(t[i].b);}cout << ans << endl;}