bin mini nes top data- n) ever current effective

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. 題意：一輛卡車要行駛L單位距離。最開始時，卡車上有P單位汽油，每向前行駛1單位距離消耗1單位汽油。如果在途中車上的汽油耗盡，卡車就無法繼續前行，即無法到達終點。途中共有N個加油站，加油站提供的油量有限，卡車的油箱無限大，無論加多少油都沒問題。給出每個加油站距離終點的距離和能夠提供的油量，問卡車從起點到終點至少要加幾次油？如果不能到達終點，輸出-1。 分析：轉換一下思考方式：在卡車開往終點的途中，只有在加油站才可以加油。但是，如果認為“在到達加油站i時，就獲得了一次在之後的任何時候都可以加Bi單位汽油的權利”，在解決問題上也是一樣的。而在之後需要加油時，就認為是在之前經過的加油站加的油就可以了。因為希望加油次數盡可能少，所以當燃料為0了之後再加油是最好的選擇。基於貪心的思想，當燃料為0時，選擇能加油量最大的加油站。所以可以用一個優先隊列來保存經過的加油站的油量，當需要加油時，取出隊列中的最大元素即可。（如此水題交了n遍，QAQ...）

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<queue>
 6 #define maxn 1000005 
 7 using namespace std;
 8 int n,l,p,a[maxn],b[maxn];
 9 struct sky{
10     int a,b;
11 }s[maxn];
12 bool cmp(sky x,sky y){return x.a<y.a;}
13 int main(){
14     ios::sync_with_stdio(false);
15     while(~scanf("%d",&n)){
16         for(int i=0;i<n;i++)cin>>s[i].a>>s[i].b;cin>>l>>p;
17         for(int i=0;i<n;i++)s[i].a=l-s[i].a;  //轉化為加油站到起點的距離
18         s[n].a=l,s[n].b=0;n++;  //把終點也認為是加油站
19         sort(s,s+n,cmp);
20         priority_queue<int> qu;
21         int pos=0,ans=0,tank=p;  //tank表示油量，pos表示位置 
22         for(int i=0;i<n;i++){
23             int d=s[i].a-pos;  //要走的距離 
24             while(tank-d<0){
25                 if(qu.empty()){puts("-1");return 0;}
26                 tank+=qu.top(); qu.pop();
27                 ans++;
28             }
29             tank-=d;
30             pos=s[i].a;
31             qu.push(s[i].b);
32         }
33         printf("%d\n",ans);
34     }
35     return 0; 
36 }

【poj 2431】Expedition