POJ2431 Expedition【贪心+优先队列】

Expedition

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16324		Accepted: 4662

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

问题链接：POJ2431 Expedition。

题意简述：

一辆卡车要行驶l单位的距离到达终点，开始时有p单位的油。途中有n个加油站，给出每个加油站到距离终点和可以加油的数量。
一个卡车的油箱无限，每走1单位距离需要消耗1单位的油，问卡车要到达终点最少需要加多少次油，如果到达不了终点则输出-1。

问题分析：

卡车从出发点开始到终点，应该按先开到距离出发点近的加油站，这样才能加更多的油，保证开更远的距离。

每当到达一个加油站，不一定就要加油。如果油够的话尽可能在油量多的加油站加油，所有把到达过的加油站可以加油的数量当作加油的权利。

油不够时，尽量用加油权利中的最大值。

这个问题用贪心法解决，包括两个方面，一是加油站由近及远，二是从大到小使用加油权利。

程序说明：优先队列默认是从大到小。

题记：（略）

AC的C++语言程序如下：

/* POJ2431 Expedition */

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>

using namespace std;

const int N = 10000;

struct _station {
    int dist, fuel;
    bool operator < (const _station& s) const {
        return dist > s.dist;
    }
} station[N];
int n, l, p;

int main()
{
    while(scanf("%d", &n) != EOF) {
        priority_queue<int> fuel;

        for(int i=0; i<n; i++)
            scanf("%d%d", &station[i].dist, &station[i].fuel);
        sort(station, station + n);     // 距离从大到小

        scanf("%d%d", &l, &p);

        fuel.push(p);
        int ans = 0, i = 0;
        while(l > 0 && !fuel.empty()) {
            ans++;
            l -= fuel.top();
            fuel.pop();
            while(i < n && l <= station[i].dist)
                fuel.push(station[i++].fuel);
        }
        printf("%d
", l <= 0 ? ans - 1: -1);
    }

    return 0;
}