POJ 2431 Expedition （贪心、优先队列）

Expedition

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7980		Accepted: 2339

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

要仔细读题 题目给的是距离终点的距离 仔细想想这道题 就是 如果当前的油量可以到达该加油站~

就存到优先队列里 然后 没有了加之前最大的就可以了 无法到达当前加油站 输出-1

AC代码如下：

//
//  POJ 2431 Expedition
//
//  Created by TaoSama on 2015-03-13
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
#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>
#define CLR(x,y) memset(x, y, sizeof(x))

using namespace std;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int N = 1e5 + 10;

int n;
pair<int, int> a[10005];

int main() {
#ifdef LOCAL
	freopen("in.txt", "r", stdin);
//	freopen("out.txt","w",stdout);
#endif
	ios_base::sync_with_stdio(0);

	while(cin >> n) {
		priority_queue<int> pq;
		for(int i = 1; i <= n; ++i)
			cin >> a[i].first >> a[i].second;
		int l, p; cin >> l >> p;
		sort(a + 1, a + 1 + n);
		a[0].first = a[0].second = 0;
		int ans = 0;
		for(int i = n; i >= 0; --i) {
			while(p - l + a[i].first < 0) {
				if(pq.empty()) {
					ans = -1;
					break;
				}
				p += pq.top(); pq.pop();
				++ ans;
			}
			if(ans == -1) break;
			pq.push(a[i].second);
		}
		cout << ans << endl;
	}
	return 0;
}