hdu 5477 A Sweet Journey 模拟

A Sweet Journey

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1070    Accepted Submission(s): 539

Problem Description

Master Di plans to take his girlfriend for a travel by bike. Their journey, which can be seen as a line segment of length L, is a road of swamps and flats. In the swamp, it takes A point strengths per meter for Master Di to ride; In the flats, Master Di will regain B point strengths per meter when riding. Master Di wonders：In the beginning, he needs to prepare how much minimum strengths. (Except riding all the time,Master Di has no other choice) 

 

Input

In the first line there is an integer t (1≤t≤50), indicating the number of test cases.
For each test case:
The first line contains four integers, n, A, B, L.
Next n lines, each line contains two integers: Li,Ri, which represents the interval [Li,Ri] is swamp.
1≤n≤100,1≤L≤105,1≤A≤10,1≤B≤10，1≤Li<Ri≤L.
Make sure intervals are not overlapped which means Ri<Li+1 for each i (1≤i<n).
Others are all flats except the swamps.

 

Output

For each text case:
Please output “Case #k: answer”(without quotes) one line, where k means the case number counting from 1, and the answer is his minimum strengths in the beginning.

 

Sample Input

12 2 2 51 23 4

 

Sample Output

Case #1: 0

 

Source

2015 ACM/ICPC Asia Regional Shanghai Online

 

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题意：一段长为L的路上有沼泽和平地，平地上每行驶单位路程恢复B点能量，沼泽上则减少A点能量。问一开始最少需要准备多少能量。

解法：关键是使得行驶到沼泽右端点时的能量大于等于0即可。于是假设一开始有0点能量，能量为负仍然可以前进。现在模拟，找到能量的最低值minE，如果>=0，那么一开始不需要准备能量，否则准备-minE点。

#include<cstdio>#include<iostream>#include<algorithm>using namespace std;const int INF =0x3f3f3f3f;int n,A,B,L;int main(){   std::ios::sync_with_stdio(false);   int T,le,ri,kase=0;cin>>T;   while(T--)   {      cin>>n>>A>>B>>L;      int p=0,e=0,mine=INF;      for(int i=1;i<=n;i++)      {          cin>>le>>ri;          e+= B*(le-p)-A*(ri-le  );          mine=min(mine,e);          p=ri;      }      int ans=mine<0?-mine:0;      printf("Case #%d: %d\n",++kase,ans);   }   return 0;}