Expedition(POJ 2431) 优先队列

来自《挑战程序设计竞赛》

1.题目原文

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

Expedition

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13194		Accepted: 3749

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.

Source

USACO 2005 U S Open Gold

2.解题思路

在卡车开往终点的路上，只有在加油站才能加油。换个思考方式，“在到达加油站i时，获得了在这次之后的任何时候都可以加B[i]单位汽油的权利”，在解决问题上也是一样的。而之后需要加油，只需要在之前经过的加油站加的油就可以了。

希望加油次数最少，所以当油量为0时加油再好不过。此外，应该选能加油量最大的加油站。

为了高效进行上述操作，我们可以使用一个从大到小的顺序依次取出数值的优先队列。

1.在经过加油站i时，往优先队列里加入B[i]。

2.当燃料箱为空时

(1).如果优先队列也是空的，无法到达终点。

(2).否则取出优先队列里的最大元素，并用来给卡车加油。

3.AC代码

#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
#include<cmath>
#include<bitset>
#include<sstream>
using namespace std;
#define INF 0x7fffffff
#define maxn 10005

//A[i]表示起点到各个加油站的距离
//B[i]表示各个加油站的可以给卡车加的汽油量
//C[i]表示各个加油站到终点的距离
int L,P,N;
int A[maxn],B[maxn],C[maxn];
pair<int,int> pir[maxn];
void solve()
{
    //为了写起来方便，我们把终点也认为是加油站
    A[N]=L;
    B[N]=0;
    N++;

    //维护加油站的优先队列
    priority_queue<int> que;

    //ans:加油次数，pos:现在所在位置，tank:油箱中汽油的量
    int ans=0,pos=0,tank=P;
    for(int i=0;i<N;i++){
        //接下来要行驶的距离
        int d=A[i]-pos;
        //不断加油直至油量足够行驶到下一个加油站
        while(tank-d<0){
            if(que.empty()){
                cout<<-1<<endl;
                return;
            }
            tank+=que.top();
            que.pop();
            ans++;
        }
        tank-=d;
        pos=A[i];
        que.push(B[i]);
    }
    printf("%d\n",ans);
}

int main()
{
    scanf("%d",&N);
    for(int i=0;i<N;i++){
        scanf("%d%d",&C[i],&B[i]);
    }
    scanf("%d%d",&L,&P);
    for(int i=0;i<N;i++){
        A[i]=L-C[i];
    }
    for(int i=0;i<N;i++){
        pir[i]=make_pair(A[i],B[i]);
    }
    //按照A[i]从小到大的顺序排序
    sort(pir,pir+N);
    for(int i=0;i<N;i++){
        A[i]=pir[i].first;
        B[i]=pir[i].second;
    }
    solve();
    return 0;
}