Expedition（优先队列）

原题链接
Expedition
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 12892 Accepted: 3682
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.

这个题乍一看好像很难去做，但其实换一种思路的话就会好做了。那就是将每一次经过的加油站保存起来，需要用的时候就把能加到的最多的油加进去，不够就继续加，如果能够走到下一个加油站就把加油站存下你继续走，如果油都加完了还走不到那说明真的是走不到了

//http://poj.org/problem?id=2431#include <algorithm>#include <iostream>#include <sstream>#include <cstring>#include <cstdio>#include <cmath>#include <queue>using namespace std;const int MOD = int(1e9) + 7;//int MOD = 99990001;const int INF = 0x3f3f3f3f;//const LL INFF = 0x3f3f3f3f3f3f3f3fLL;//const DB EPS = 1e-9;//const DB OO = 1e20;//const DB PI = acos(-1.0); //M_PI;const int fx[] = {-1, 1, 0, 0};const int fy[] = {0, 0, -1, 1};const int maxn=10000 + 10;typedef struct node{        int loc,fuel;}node;node a[maxn];int n,L,P;int cmp(node x,node y){        return x.loc < y.loc;}void solve(){        priority_queue<int> q;        //ans：加油的次数 pos：现在所在的位置 tank：油箱中还剩的油        int ans=0,pos=0,tank=P;        for(int i=0;i<n;i++){                int d=a[i].loc-pos;//还要前进的距离                while(tank-d<0){//油不够就加油                        if(q.empty()){                                puts("-1");                                return;                        }                        else{                                tank+=q.top();                                q.pop();                                ans++;                        }                }                tank-=d;//能到那么油箱中就要把这些路程的油烧掉                pos=a[i].loc;//到达新位置                q.push(a[i].fuel);        }        printf("%d\n",ans);}int main(){        cin >> n;        for(int i=0;i<n;i++)                scanf("%d%d",&a[i].loc,&a[i].fuel);        cin >> L >> P;        for(int i=0;i<n;i++)                a[i].loc=L-a[i].loc;        a[n].loc=L;//为了便于处理，我们把终点也认为是加油站        a[n].fuel=0;        n++;        sort(a,a+n,cmp);        solve();        return 0;}