A Sweet Journey

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本文介绍了一个名为ASweetJourney的问题，主人公计划骑自行车带着女友旅行，需要计算在特定地形条件下所需的最小初始体力值。地形包括湿地和平原，分别对应不同的体力消耗和恢复规则。

A Sweet Journey

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

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int a[100010];
int main()
{
    int T;
    cin>>T;
    for(int k=1; k<=T; k++)
    {
        memset(a,0,sizeof(a));//0代表休息，1代表耗费力气
        int n,x,y,l;
        int zuo,you;
        scanf("%d%d%d%d",&n,&x,&y,&l);
        for(int i=0; i<n; i++)
        {
            scanf("%d%d",&zuo,&you);
            for(int j=zuo; j<=you-1; j++)//将区间简化为左端点记录在数组中；
            {
                a[j]=1;
            }
        }
        int po=0,min=y;
        for(int i=0; i<l; i++)
        {
            if(a[i]==0)
                po+=y;//休息，加体力；
            if(a[i]==1)
                po-=x;//消耗体力；
            if(min>po)
                min=po;//找出最小临界点即可；
        }
        printf("Case #%d: ",k);
        if(min>=0)
            printf("0\n");//最小为正就不需要带体力值；
        if(min<0)
            printf("%d\n",-min);
    }
    return 0;
}