Expedition
Time Limit: 1000MS
Memory Limit: 65536K
Total Submissions: 9148
Accepted: 2673
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.
這是挑戰程序設計競賽上的關於 堆 以及 優先隊列運用的例題，書上給出的思路非常巧妙
因為求得是最優解，需要盡可能少的加油站，所以我們每次希望去加油的時候 加最大的那個，因而將加油站push進priority_queue(堆結構，默認每次彈出最大值)
從起點開始跑，把路程通過加油站分成幾部分，每一部分只要通過加油能到達加油站即可，，因而，把終點也視為加油站 距離L，加油量 0；
#include //輸入量較大，所以用c風格來輸入輸出， ac時間 16ms ， 全部換成c++的風格則需要 313ms
#include
#include
using namespace std;
struct node{
int a1,b1;
};
int cmp(const node& a,const node& b){
return a.a1 pq;
int n;
node no[10010];
while(scanf(%d,&n)!=EOF){
for(int i=0;i
下面是另一種方法
#include
#include
#include
using namespace std;
struct node{
int a1,b1;
}no[10010];
int cmp(const node&a, const node&b){
return a.a1pq;
while(scanf(%d,&t)!=EOF){
while(!pq.empty())
pq.pop();
for(int i=0;i=l) //當最遠跑的距離超過終點
printf(%d
,cnt);
else
printf(-1
);
}
return 0;
}
