HDU 5477A Sweet Journey网赛

A Sweet Journey

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 175    Accepted Submission(s): 87

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
   
   
   
   

 
题意是,在一个L长的道路上 存在减少能量的的地方存在, 剩余地方都是增加能量的地方
问初始的能量应该为多少,才能使人安全通过.
即在行驶过程中不能使能量和为负数了

Source

2015 ACM/ICPC Asia Regional Shanghai Online

 

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#include<iostream>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#define INF 0x3f3f3f3f

using namespace std;

int vis[600000];
int main()
{
	int T,n,A,B,L,a,b;
	while(~scanf("%d",&T))
	{
		int Case=0;
		while(T--)
		{
			Case++;
			scanf("%d%d%d%d",&n,&A,&B,&L);
			for(int i=0;i<L;i++)
			  vis[i]=B;
			for(int i=0;i<n;i++)
			{
				scanf("%d%d",&a,&b);
				for(int j=a;j<b;j++)
					vis[j]=-A;
			}
			long long sum=0,Min=0;
          //int sum=0;
			for(int i=0;i<=L;i++)
			{
				sum+=vis[i];
//				cout<<sum;
//				cout<<sum;
				if(sum<Min)
					Min=sum;
			}
           //cout<<Min;
           printf("Case #%d: ",Case);
		    printf("%lld\n",0-Min);
		}
	}
}

 

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