贪心，优先级队列。

基本思路：走过一个加油站，先不要加油，把这个油量存到仓库，到油量不够的时候去仓库补油，补油优先选择油量大的。

细节较多，容易写错。

#include
     
      #include
      
       #include
       
        #include
        
         #include
         
          using namespace std;int n;struct Point{    long long x;    long long sum;}p[100000 + 10];long long L, P;bool cmp(const Point&a, const Point&b){    return a.x
          

q; for (int i = 1; i <= n; i++) scanf("%lld%lld", &p[i].x, &p[i].sum); scanf("%lld%lld", &L, &P); for (int i = 1; i <= n; i++) p[i].x = L - p[i].x; sort(p + 1, p + 1 + n, cmp); if (p[n].x != L){ p[n + 1].x = L; p[n + 1].sum = 0; n++; } int ans = 0; long long pre = 0; for (int i = 1; i <= n; i++) { long long dis = p[i].x - pre; P = P - dis; if (p[i].x == L&&P >= 0) break; if (p[i].x == L) { while (P < 0 && (!q.empty())) { P = P + q.top(); q.pop(); ans++; } if (P < 0) ans = -1; break; } else { pre = p[i].x; if (P == 0) { q.push(p[i].sum); P = P + q.top(); q.pop(); ans++; } else if (P < 0) { while (P < 0 && (!q.empty())) { P = P + q.top(); q.pop(); ans++; } if (P < 0) { ans = -1; break; } else if (P == 0) { q.push(p[i].sum); P = P + q.top(); q.pop(); ans++; } else if (P>0) q.push(p[i].sum); } else q.push(p[i].sum); } } printf("%d\n", ans); } return 0;}