poj2431 Expedition优先队列

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.

 

这到题的大意是是一个卡车路经许多加油站，每个加油站可加的油有上限，问到达终点最少需加几次油。

做这道题思维要变化一下，我们假设没经过一个加油站就获得了一张随时可变成油的魔法卡片。

当我们发现没油时，就去使用一张卡片，此时当然需要用那张油量最多的卡片。

 

使用priority_queue需#include <queue>

using namespace std

方法有push(), top(), pop(), empty()

它会每回pop()一个最大值出来

 

代码如下

 #include <cstdio>
 #include <iostream>
 #include <algorithm>
 #include <cstring>
 #include <queue>

 using namespace std;
 int n;
 int l, p;
 struct Fuel{
   int disD;
   int disS;
   int c;
 };
 Fuel fuel[];

 int cmp(const void *a, const void *b) {
   Fuel at = *(Fuel *)a;
   Fuel bt = *(Fuel *)b;
   return at.disS - bt.disS;
 }

 priority_queue <int> fque;

 int main(int argc, char const *argv[])
 {
     while(scanf("%d",&n) != EOF) {
         for(int i = ; i < n; i++) {
             int tmp;
             scanf("%d %d",&fuel[i].disD,&fuel[i].c);
         }
         scanf("%d %d",&l, &p);
         for(int i = ; i < n; i++) {
             fuel[i].disS = l - fuel[i].disD;
         }
         fuel[n].disS = l;
         fuel[n].c = ;
         qsort(fuel,n, sizeof(Fuel), cmp);

         while(!fque.empty()) {
               fque.pop();
         }
         int ans = ;
         int now = ;
         int flag = true;
         for(int i = ; i <= n && flag; i++) {
             int remain = p - fuel[i].disS + now;
             while(remain < ) {
                 if(fque.empty()) {
                     flag = false;
                     break;
                 }
                 int tmp = fque.top();
                 fque.pop();
                 p = p + tmp;
                 remain = remain + tmp;
                 ans++;
             }
             if(remain >= ) {
                 fque.push(fuel[i].c);
                 now = fuel[i].disS;
                 p = remain;
             } 

         }
         if(now == l) {
             printf("%d\n",ans);
         }
         else {
           puts("-1");
         }

     }
     return ;
 }