poj2431 Expedition 贪心+优先队列 【挑战程序设计竞赛】

题目链接http://poj.org/problem?id=2431

Expedition

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 28629 Accepted: 7935

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

思路

数据预处理：将题目中离终点的距离，改成离起点的距离
数据存储：以结构体形式保存到vector中，并以离起点的距离从小到大排序。对于路经的加油站，放入到优先队列中，油量大的加油站优先
处理过程：
1.采用贪心策略，如果车此时最大能行驶距离P还不能到达下一个加油站，就在P距离之前的加油站选一个加油，然后P就变大了，然后再继续走，又遇到不能到达下一个加油站的时候，就在之前的加油站再次加油，若此时队列为空，说明就无法加油，加油站都走不完肯定也到达不了终点，return -1,。
2.当走完了所有加油站，然后再判断当前的P是否可以到达终点，不能的话继续在队列中加油，如果队列取到空，还没到达终点 return -1
3.经过上面2步，没有return -1，就返回加油的次数

AC代码

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define PI acos(-1)
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const ll maxN = 1e6+10; 

struct node{
	int pos,oil;
	friend bool operator < (node n1,node n2){
		return n1.oil < n2.oil; //大根堆 
	}
};

bool cmp(node n1,node n2){
	return n1.pos<n2.pos;
}
ll n,L,P;
vector<node> ve;
priority_queue<node> q; 

int fun(){ //P:当前能到达的距离 
	int t = 0;//加油次数 
	for(int i = 0;i<ve.size();){
		if(ve[i].pos <= P){ //将此时能到达距离内的加油站放入优先队列 
			q.push(ve[i]);
			i++;
		}else{  
			if(q.empty()) return -1; //到达不了下一个加油站，又不能加油，说明不能到达终点 
			P+=q.top().oil;
			q.pop();
			t++;
		}
	}
	while(P<L){ //经过上面for循环的加油，还不能到达终点，就尝试是否可以通过继续加油到达终点 
		if(q.empty()) return -1;
		P+=q.top().oil;
		q.pop();
		t++;
	}
	return t;
}

int main(){
	cin>>n;
	int t1,t2;
	for(int i = 0;i<n;i++){
		scanf("%d%d",&t1,&t2);
		ve.push_back({t1,t2});
	}
	cin>>L>>P; 
	for(int i = 0;i<ve.size();i++){ //将离终点的距离，改成离起点的距离 
		ve[i].pos= L-ve[i].pos;
	}
	sort(ve.begin(),ve.end(),cmp);//按离起点从小到大排序 
	cout<<fun()<<endl;
	return 0;
}