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POJ 2431 Expedition (优先队列)

题目

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 $N$ $(1 \leq N \leq 10,000)$ 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 $(1 \leq P \leq 1,000,000)$ .

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

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4
4 4
5 2
11 5
15 10
25 10

Sample Output

1
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.

Source

USACO 2005 U S Open Gold


题意

你需要驾驶一辆卡车行驶 $L$ 单位距离。最开始时,卡车上有 $P$ 单位的汽油。卡车每开 $1$ 单位距离就要消耗 $1$ 单位汽油。如果在途中车上的汽油耗尽,卡车就无法继续前行,因而无法到达终点。在途中一共有 $N$ 个加油站。第 $i$ 个加油站在距离起点 $A_i$单位距离的地方,最多可以给卡车加 $B_i$单位汽油。假设卡车的燃料箱的容量是无限大的,无论加多少油都没有问题。那么问卡车是否能到达终点?如果可以,最少需要加多少次油?如果不能,输出 $-1$ 。


分析

由于加油站数量非常大,所以算法必须高效。这样想,在卡车开往终点的途中,只有在加油站才可以加油,但是,如果认为有只要经过这个加油站在任何时候都可以加 $B_i$ 单位汽油的权利 ,所以在之后需要加油的时候,就认为是在之前经过的加油站加的油就可以了。那么,当燃料为 $0$ 时应该用哪个加油站加油呢,因为油箱的容量无限大,所以挑选在此时位置之前所经过的加油站油量最大的加油站。

为了高效地进行上述操作,我们可以使用从大到小地顺序依次取出数值地优先队列

上面的分析是《挑战算法竞赛》的一些分析,算法的确巧妙,但是自己想的确有点吃力,所以照着书本的分析,写出了代码,但是由于树上的代码是为了方便只写了关键代码,而且“为了方便起见,将题目输入的加油站到终点的距离改成了起点到加油站的距离” ,这里 $WA$ 了不少次,然后改对之后,是由于题目输入的数据不是按照加油站距离起点远近顺序,因此得需要建立结构体按照加油站与起点的距离排序。


代码

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#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
#include<stack>
#include<vector>
#include<cmath>
#include<set>
#include<map>
#include<cstdlib>
#include<functional>
#include<climits>
#include<cctype>
#include<iomanip>
using namespace std;
typedef long long ll;
#define INF 0x3f3f3f3f
#define mod 1e9+7
#define clr(a,x) memset(a,x,sizeof(a))
const double eps = 1e-6;
const int MAXN=1000000;
int l,p,n;
struct node
{
int a,b;
}s[MAXN];
bool cmp(const node c,const node b)
{
return c.a<b.a;
}
void solve()
{
for(int i=0;i<n;i++)
{
s[i].a=l-s[i].a; //转换为加油站到起点的距离
}
s[n].a=l;
s[n].b=0;
n++;
sort(s,s+n,cmp);
int ans=0; //加油次数
int pos=0; //现在所在位置
int tank=p; //油箱中的油量
priority_queue<int> que;
for(int i=0;i<n;i++)
{
int d=s[i].a-pos; //当前需要走的距离
while(tank-d<0)
{
if(que.empty()) //油箱为空
{
puts("-1");
return;
}
else
{
tank+=que.top();
que.pop();
ans++;
}
}
tank-=d;
pos=s[i].a;
que.push(s[i].b);
}
printf("%d\n",ans);
}
int main()
{
while(scanf("%d",&n)!=EOF)
{
for(int i=0;i<n;i++)
{
scanf("%d%d",&s[i].a,&s[i].b);
}
scanf("%d%d",&l,&p);
solve();
}
return 0;
}
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