HDU5477 A Sweet Journey

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 (), 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: , which represents the interval  is swamp. 
，. 
Make sure intervals are not overlapped which means  for each i (). 
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

如果把这个看成赚钱和还钱的话，骑在平地上是赚了，骑在沼泽是亏了，开始本金是0，然后就找在骑得过程中最多亏多少，那就是我们要找的最小的所需要的体力。

#include <stdio.h>int main(){int t,icase;scanf("%d",&t);for(icase=1;icase<=t;icase++){int n,A,B,L;int i,j,k;scanf("%d%d%d%d",&n,&A,&B,&L);int min=0,power=0;int point=0;for(i=0;i<n;i++){int r,l;scanf("%d%d",&r,&l);power-=(r-point)*B;power+=(l-r)*A;if(power>min)min=power;point=l;}printf("Case #%d: %d\n",icase,min);}return 0;}