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

4
4 4
5 2
11 5
15 10
25 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：

给出一辆车初始的油量P和距离城镇的距离L，然后给出距离城镇的加油站的距离和可加油量，求要是汽车到达城镇最少停留加油的次数，然后车的油箱容量是无限的。

思路：

很标准的一道贪心题，但是和之前的题有所不同，排序方式不同，当油量为0再加油是最好的选择，但是并不是每次没油了都能刚好遇到加油站，换一种思路就是找在车没油的这段路程中能加油最多的加油站，就在这个加油站加油。

自己的代码写的太复杂，去看了一下大佬的代码，简单很多

代码如下

#include<bits/stdc++.h>
using namespace std;
int L,P,N;
int A[10010],B[10010];
void init()
{
    cin>>N;
    for(int i=0;i<N;i++)
       cin>>A[i]>>B[i];
    cin>>l>>p;
    A[N]=L; //为了计算方便，将终点也设为加油站
    B[N]=0;
}
int solve()
{
    priority_queue<int> que;
    int ans=0;//加油次数
    int pos=0;//所在位置
    int tank=P;//剩余油量
    for(int i=0;i<N;i++)
        {
        int d=A[i]-pos;
       while (d>tank)
       {
           if(que.empty())
           {
              ans=-1;
               return ans;
           }
           tank+=que.top();
           que.pop();
           ans++;
       }
       tank=tank-(A[i]-pos);
       pos=A[i];
       que.push(B[i]);
    }
    return ans;
}
int main()
{
    init();//输入
    cout<<solve();//输出
    return 0;
}