hdu5477 A Sweet Journey

A Sweet Journey

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 504 Accepted Submission(s): 279

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：在区间 [0, L] 中有沼泽和陆地，经过沼泽每米损失a体力，经过陆地每米回复b体力，要从0到L；求在0处所需的最小体力。

分析：因为沼泽都是按顺序排列的，只要把每个沼泽处理一遍，找出某个时候需要的损失的最大体力。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
#define ll long long
#define CL(a) memset(a,0,sizeof(a))

int main ()
{
    int T,n,a,b,L;
    cin>>T;
    for (int cas=1; cas<=T; cas++)
    {
        cin>>n>>a>>b>>L;
        int p=0,sum=0,minx=99999;//p用来标记上一个沼泽结束位置
        int l,r;
        while (n--)
        {
            cin>>l>>r;
            sum = sum+b*(l-p)-a*(r-l);//没经过一片沼泽所剩于体力
            minx = min(minx, sum);//途中所需的最大体力
            p = r;//更新p
        }
        cout<<"Case #"<<cas<<": ";
        if (minx < 0)
            cout<<-minx<<endl;
        else
            cout<<"0"<<endl;
    }
    return 0;
}