ACM--过沼泽--模拟--HDOJ 5477--A Sweet Journey

HDOJ题目地址：传送门

A Sweet Journey

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 759    Accepted Submission(s): 397

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

   
   
   
   

    
    
    
    1
2 2 2 5
1 2
3 4
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case #1: 0
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   

题意：一个人去旅行，路上有沼泽和平路，当走平路是增加a点体力，当走沼泽时消耗b点体力，求最开始最少要携带多少点体力

#include<iostream>
#include<stdio.h>
#include<memory.h>
#include<algorithm>
using namespace std;
struct Node{
   int begin;
   int end;
   int chazhi;
}zhaoze[101];
bool cmp(Node a,Node b){
   if(a.begin<b.begin)
	   return true;
   return false;
}
int main(){
  int n,m,i,a,b,l,kaishi,jieshu,result,temp,index=1;
  cin>>n;
  while(n--){
      cin>>m>>a>>b>>l;
	  for(i=0;i<m;i++){
	     cin>>zhaoze[i].begin>>zhaoze[i].end;
		 zhaoze[i].chazhi=zhaoze[i].end-zhaoze[i].begin;
	  }
	  sort(zhaoze,zhaoze+m,cmp);
	  result=0;
	  temp=0;
	  kaishi=0;
	  jieshu=0;
	  for(i=0;i<m;i++){
	     temp=temp+(zhaoze[i].begin-jieshu)*b-zhaoze[i].chazhi*a;
		 jieshu=zhaoze[i].end;
		 if(result>temp){
		   result=temp;
		 }
	  }
	  temp+=(l-jieshu)*b;
	  if(result>temp){
		   result=temp;
	  }
	  if(result<0){
	       printf("Case #%d: %d\n",index,-result);
	  }else{
	      printf("Case #%d: 0\n",index);
	  }
	  index++;
  }
}