HDU 5477 A Sweet Journey

A Sweet Journey

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

Total Submission(s): 703    Accepted Submission(s): 363

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
   
   
   
   

 

Source

2015 ACM/ICPC Asia Regional Shanghai Online 

题意：一个人骑自行车旅行，路的长度为L，陆地上每里获得b点力量，沼泽中每里失去a点力量，n为沼泽的数量，下面的nz组数据代表沼泽的起点和终点。

沼泽的顺序都是从起点到终点的，不用考虑顺序。简单模拟接可以了。

AC代码：

#include <cstdio>
#include <algorithm>
#include <algorithm>
using namespace std;
const int INF = 0x3f3f3f3f;
int main()
{
    int t,a,b,l,n;
    int kase=0;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d%d",&n,&a,&b,&l);
        int sum=0;
        int last=0;
        int maxn=INF;
        while(n--)
        {
            int start,end;
            scanf("%d%d",&start,&end);
            sum+=(b*(start-last)-a*(end-start));
            maxn=min(maxn,sum);
            last=end;
        }
        printf("Case #%d: ",++kase);
        if(maxn<0)
            printf("%d\n",-maxn);
        else
            printf("0\n");
    }
    return 0;
}