POJ 2431Expedition

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.

 

挑战程序设计上边的例题，自己傻，搞了1个多小时，然后还是瞎胡写。

在卡车开往终点的途中，只有在加油站才可以加油。但是，如果认为“在到达加油站 i 时，就获得了一次在任何时候就能加 Bi 单位汽油的权利”。在解决问题上是一样的。经过这样的变通之后，在需要加油的时候，认为在之前的加油站里已经加过油了。

题目中给的是每个加油站距离终点的距离，我们转换一下，让 Ai 是加油站距起点的距离。

在经过加油站时，往优先队列里加入Bi[]

当油箱空了

　　如果优先队列也是空的，则到达不了终点

　　否则，取出优先队列里最大的元素，给卡车加油。

 #include <cstdio>
 #include <algorithm>
 #include <queue>
 
 using namespace std;
 const int MAXN =  + ;
 int L, P, N;
 struct node{
     int a, b;
 }p[MAXN];
 
 bool cmp(const struct node p1, const struct node p2)
 {
     return p1.a < p2.a;
 }
 
 void solve()
 {
     p[N].a = L;
     p[N].b = ;
     N++;
 
     priority_queue<int> que;
     int pos = , tank = P, ans = ;
     for(int i = ; i != N; ++i)
     {
         int d = p[i].a - pos;
         while(tank - d < )
         {
             if(que.empty())
             {
                 puts("-1");
                 return;
             }
             tank += que.top();
             que.pop();
             ans++;
         }
         tank -= d;
         pos = p[i].a;
         que.push(p[i].b);
     }
     printf("%d\n", ans);
 }
 
 int main()
 {
     //freopen("in.txt", "r", stdin);
     while(~scanf("%d" ,&N))
     {
        for(int i = ; i != N; ++i)
        {
            scanf("%d %d", &(p[i].a), &(p[i].b));
            //printf("%d %d\n", p[i].a, p[i].b);
        }
 
        scanf("%d %d", &L, &P);
         for(int i = ; i != N; ++i)
             p[i].a = L - p[i].a;
         sort(p, p+N, cmp);
 
         solve();
     }
     return ;
 }