poj 2431Expedition（优先队列+贪心）

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

44 45 211 515 1025 10

Sample Output

2

solution：

这道题是明显的贪心问题，我们可以把每经过一个加油站i，认为拥有加b[i]这么多汽油的权利，当当前油量不足以开到当前地点，那么就加油，直到可以到达该点。

#include<cstdio>#include<queue>#include<algorithm>using namespace std;const int maxn = 1e5 + 200;priority_queue<int>q;struct node{int dis, v;}a[maxn];bool cmp(node a, node b)//题目中未标明距离是按顺序给的{return a.dis < b.dis;}int main(){int n, l, p;while (~scanf("%d", &n)){for (int i = 0; i < n; i++)scanf("%d%d", &a[i].dis, &a[i].v);scanf("%d%d", &l, &p);for (int i = 0; i < n; i++)a[i].dis = l - a[i].dis;sort(a, a + n, cmp);a[n].dis = l; a[n].v = 0;//为了方便，我们把终点看做一个加油站int ans = 0, pos = 0, tank = p, f = 0;//pos为当前的位置，tank为当前的油量for (int i = 0; i <=n; i++){int d = a[i].dis - pos;//d为当前要走的距离while (tank < d){if (q.empty())f = 1;ans++;tank += q.top();q.pop();}tank -= d;pos = a[i].dis;q.push(a[i].v);}if (f)printf("-1\n");else  printf("%d\n", ans);}}