poj 2431 【优先队列】

poj 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. 

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：一辆卡车从起点行驶到终点路程为L，每行驶单位距离花费单位油量。起始时有油量P。在路途中有N个加油站，距离终点距离为Ai,有油量Bi。车如果没油就不能到达终点，假设车油箱无限大。问最少加几次油才能到达终点，不能达到输出-1。

题解：把加油站距离转化为到起点的距离便于操作。可以把问题看成在经过第i个加油站时，获得了加Bi油量的权利，等以后需要加油了就直接加。显然为了使加油次数最少，需要每次选尽可能大的Bi加油，所以需要优先队列。为方便操作，把终点也看成一个加油站。

 

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
struct No{
    int A;
    int B;
};
No C[10050];
int N,P,L;

bool cmp(No a,No b)
{
    return a.A>b.A;
}

int main()
{
    int ans=0;
    while(cin>>N)
    {
        for(int i=0;i<N;i++) cin>>C[i].A>>C[i].B;
        cin>>L>>P;
        sort(C,C+N,cmp);N++;
        C[N].A=L,C[N].B=0;
        int pos=0,tank=P;
        priority_queue<int> que;
        for(int i=0;i<N;i++)
        {
            C[i].A=L-C[i].A;
            int d=C[i].A-pos;
            while(tank-d<0)
            {
                if(que.empty()){
                    cout<<"-1"<<endl;
                    return 0;
                }
                tank+=que.top();
                que.pop();
                ans++;
            }
            tank-=d;
            pos=C[i].A;
            que.push(C[i].B);
        }
        cout<<ans<<endl;
    }
    return 0;
}