POJ 2431 Expedition 贪心简单的优先级队列问题

刺骨的言语ヽ痛彻心扉 2023-07-12 05:05 21阅读 0赞

Expedition

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;65536K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;30702</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;8457</td> 
  </tr> 
 </tbody> 
</table>

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

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.

Source

[USACO 2005 U S Open Gold][]

1 #include <cstdio>
     2 #include <iostream>
     3 #include <queue>
     4 #include <algorithm>
     5 
     6 using namespace std;
     7 
     8 const int max_n = 1e4+2;
     9 const int max_L = 1e6;
    10 const int max_P = 1e6;
    11 const int max_A = max_L;
    12 const int max_B = 100;
    13 
    14 int n,L,P;
    15 int a[max_n],b[max_n];
    16 
    17 typedef struct Node
    18 {
    19     int a,b;
    20 };
    21 Node node[max_n];
    22 
    23 
    24 bool cmp(Node a,Node b)
    25 {
    26     return a.a<b.a;
    27 }
    28 
    29 void solve()
    30 {
    31     // 为了方便起见，将终点看作加油站
    32     node[n].a=L;
    33     node[n].b=0;
    34     ++n;
    35 
    36     // 对数组按照距离排序
    37     // 白书害我，我以为不需要排序，疯狂wa
    38     sort(node,node+n,cmp);
    39 
    40 //    for(int i=0;i<n;++i)
    41 //    {
    42 //        cout<<node[i].a<<' '<<node[i].b<<endl;
    43 //    }
    44 
    45     int ans=0;
    46 
    47     // 定义最大优先级队列，存储可以到达位置的油量
    48     priority_queue<int> heap;
    49 
    50     for(int i=0;i<n;++i)
    51     {
    52         // 当前可达终点，跳出循环
    53         if(P>=L)
    54         {
    55             break;
    56         }
    57         // 当前点不可达
    58         while(P<node[i].a)
    59         {
    60             // 加油，直到可达或者为空
    61             if(heap.empty())
    62             {
    63                 puts("-1");
    64                 return;
    65             }
    66             ++ans;
    67             P+=heap.top();
    68             heap.pop();
    69         }
    70 
    71         // 当前点可达，将当前点加入堆
    72         heap.push(node[i].b);
    73     }
    74 
    75     printf("%d\n",ans);
    76 }
    77 
    78 int main()
    79 {
    80     scanf("%d",&n);
    81     int dis,fuel;
    82     for(int i=0;i<n;++i)
    83     {
    84         scanf("%d %d",&dis,&fuel);
    85         node[i].a=dis;
    86         node[i].b=fuel;
    87     }
    88 
    89     scanf("%d %d",&L,&P);
    90 
    91     for(int i=0;i<n;++i)
    92     {
    93         node[i].a=L-node[i].a;
    94     }
    95     solve();
    96     return 0;
    97 }

[USACO 2005 U S Open Gold]: http://poj.org/searchproblem?field=source&key=USACO+2005+U+S+Open+Gold