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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

中文意思：辆卡车要行驶L单位距离。最开始时，卡车上有P单位汽油，每向前行驶1单位距离消耗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，即无法到达终点。途中共有N个加油站，加油站提供的油量有限，卡车的油箱无限大，无论加多少油都没问题。给出每个加油站距离终点的距离和能够提供的油量，问卡车从起点到终点至少要加几次油？如果不能到达终点，输出-1。

解：优先队列+贪心；
每过一个加油站，就拥有了这个加油站的加油权利，当什么时候油不够时，就将经过的加油站中挑出最大的进行加油。
AC代码：

include

include

include

using namespace std;
pair<int ,int > a[10005];
int cmp(pair<int ,int > &a,pair<int ,int > &b){
return a.first<b.first;
}
int main()
{
int n,b;
while(cin>>n){
priority_queue<int,vector,less >q;
for(int i=1;i<=n;i++)
scanf("%d%d",&a[i].first,&a[i].second);
int k,f;
cin>>k>>f;
for(int i=1;i<=n;i++)
a[i].first=k-a[i].first;
a[0].first=0;
sort(a,a+n+1,cmp);
int s=f,sum=0,plug=0;
a[n+1].first=k,a[n+1].second=0;
for(int i=1;i<=n+1;i++){
if(s>=a[i].first) q.push(a[i].second);
else{
int plu=1;
while(!q.empty()){
int c=q.top();s+=c;q.pop();sum++;
if(s>=a[i].first){
plu=0;
q.push(a[i].second);
break;
}
}
if(plu) {
cout<<"-1"<<endl;
plug=1;
break;
}
}
}
if(!plug) cout<<sum<<endl;
}
return 0;
}

注意：sort是左闭右开，在这里错了一个小时了。。。。。

poj2431Expedition

标签：integer   mit   city   mes   term   make   无限   add   run   

原文地址：https://www.cnblogs.com/sunjianzhao/p/11394753.html