POJ2431 Expedition

Source

http://poj.org/problem?id=2431

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.

题意:你开一辆卡车要走L长的路,中间右若干加油站,每个加油站的加油量不同,求到终点的最少加油次数。

思路:《挑战程序设计》p75的例题,变换一下思考方式,如果认为“在到达加油站i时,就获得了一次在之后的任何时候都可以加B(i)单位油的权利”,就很好解决了,用优先队列,把之前到的加油站全放进去,当当前没油时,取最大的加上。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
#define maxn 100000+100
#define ll long long

int n,d[maxn],gas[maxn],r[maxn];
priority_queue<int> Q;

bool cmp(int a,int b)
{
	return d[a]<d[b];
}
int main()	
{
	//freopen("input.in","r",stdin);
	cin>>n;
	for(int i=1;i<=n+1;i++)r[i]=i;
	for(int i=1;i<=n+1;i++)cin>>d[i]>>gas[i];
	sort(r+1,r+1+n,cmp);
	int ans=0,tank=gas[r[n+1]];
	for(int i=n;i>=0;i--)
	{
		while(tank-(d[r[i+1]]-d[r[i]])<0)
		{
			if(Q.empty()){puts("-1");exit(0);}
			ans++;tank+=Q.top();Q.pop();
		}
		tank-=(d[r[i+1]]-d[r[i]]);
		Q.push(gas[r[i]]);
	}
	cout<<ans<<endl;
	return 0;
}