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：
一辆卡车要行驶L单位距离，卡车上有P单位的汽油。一共有N个加油站，分别给出加油站距终点距离，及加油站可以加的油量。问卡车能否到达终点？若不能输出-1，否则求出最少加油次数。
分析：
因为要计算出的是最少的加油数目，所以应该选择的是加油量最多的加油站，可以认为"在到达加油站之后，获得一次可以在任何时候使用这个加油站加油的资格"，因此在之后需要加油时，就认为是在之前经过的加油站加的油就可以了。因为希望加油次数最少，所以在燃料不够行驶时选择加油量最大的加油站加油。为了高效性，我们可以使用从大到小的顺序依次取出数值的优先队列。
代码：
#include<iostream>
#include<stdio.h>
#include<queue>
#include<algorithm>
using namespace std;
int n,L,p;
struct Node
{
int dis;///距离
int val;///能加的油量
} node[10009];
bool cmp(Node a,Node b)///结构体按照距离从小到大排序
{
return a.dis<b.dis;
}
int main()
{
scanf("%d",&n);
for(int i=0; i<n; i++)
scanf("%d%d",&node[i].dis,&node[i].val);
scanf("%d%d",&L,&p);
///把终点也看做是一个加油站，只不过加油量是0
node[n].dis=L;
node[n].val=0;
///因为题目上给出的是加油站到重点的距离，所以应该吧它们转换成到起点的距离
for(int i=0; i<n; i++)
{
node[i].dis=L-node[i].dis;
}
sort(node,node+n,cmp);
priority_queue<int>qu;///优先队列中，应该先选择加油量比较大的
int ans=0,start=0,soil=p;
for(int i=0; i<=n; i++)
{
int d=node[i].dis-start;///需要走的路程
while(soil<d)///当油量不足以走这些路程的时候，就要选择加油站加油
{
if(qu.empty())///优先队列为空时，不能加油了
{
printf("-1\n");
return 0;
}
ans++;
soil+=qu.top();
qu.pop();
}
///走过这段路后，就要把这个加油站的油量加入队列中
soil-=d;
qu.push(node[i].val);///起始点也要更新
start=node[i].dis;
}
printf("%d\n",ans);
return 0;
}
