简单的说说思路，如果一开始能够去到目的地那么当然不需要加油，否则肯定选择能够够着的油量最大的加油站加油，,不断重复这个贪心的策略即可。

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <queue>

using namespace std;

const int maxn=1e4+9;

int dist,p,n;

struct S

{

    int d,f;

    bool operator <(const S & xx) const

    {

        return d<xx.d;

    }

}stop[maxn];

struct cmp

{

    bool operator ()(const S &a,const S &b) const

    {

        return a.f<b.f;

    }

};

int main()

{

    while(scanf("%d",&n)!=EOF)

    {

        for(int i=1;i<=n;i++)

        scanf("%d %d",&stop[i].d,&stop[i].f);

        scanf("%d %d",&dist,&p);

        for(int i=1;i<=n;i++)

        stop[i].d=dist-stop[i].d;

        sort(stop+1,stop+1+n);

        int ans=0,t=1;

        bool flag=true;

        priority_queue <S,vector<S>,cmp> q;

        while(p<dist)

        {

            while(t<=n&&stop[t].d<=p) q.push(stop[t++]);

            if(q.empty())

            {

                flag=false;

                break;

            }

            ans++;

            p+=q.top().f;

            q.pop();

        }

        if(flag) cout<<ans<<endl;

        else cout<<-1<<endl;

    }

    return 0;

}