Working in Beijing

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others) Total Submission(s): 2113    Accepted Submission(s): 901  

Problem Description

Mr. M is an undergraduate student of FDU. He finds an intern position in Beijing, so that he cannot attend all the college activities. But in some conditions, he must come back to Shanghai on certain date. We can assume the important activities that Mr. M must attend are occupy a whole day. Mr. M must take flight to Shanghai before that day and leave after that day. On the other hand, Mr. M is absent in Beijing and he will lose his salary for his absent. Sometimes the cost of flight is much higher than the loss of salary, so to save the cost on the travel, Mr. M can stay in Shanghai to wait for another important date before he back to Beijing. Now, Mr. M knows all of the important date in the next year. Help him schedule his travel to optimize the cost.

 

Input

The input contains several test cases. The first line of single integer indicates the number of test cases.   For each test case, the first line contains three integers: n, a and b, denoting the number of important events, the cost of a single flight from Beijing to Shanghai or Shanghai to Beijing and the salary for a single day stay in Beijing. (1 <= n <= 100000, 1 <= a <= 1000000000, 1 <= b <=100)   Next line contains n integers ti, denoting the time of the important events. You can assume the ti are in increasing order and they are different from each other. (0 <= ti <= 10000000)

 

Output

For each test case, output a single integer indicating the minimum cost for this year.

 

Sample Input

2

1 10 10

5

5 10 2

5 10 15 65 70

 

Sample Output

Case #1: 30

Case #2: 74

 

Source

 

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讀題理解題意花了很久...其實就很簡單比較一下，我們居然想了很久！

【題意】 

輸入n,a,b，n是放假天數（放假不扣錢），a:單程機票錢，b：曠工扣的工資錢

如果兩個假期的間隔天數曠工扣的工資錢（b*間隔天數） < 先回去，再來的往返機票錢（2*a），那妥妥的不回去划算呀～～

所以比較下就好了，算下最小花費

【AC程式碼】

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <iostream>
#define go(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
#define ll long long
const int maxn = 1e5 + 5;
ll num[maxn];
int main()
{
    ll t,n,a,b,sum;
    scanf("%lld",&t);
    go(i,1,t)
    {
        sum = 0;
        scanf("%lld%lld%lld",&n,&a,&b);
        go(j,0,n-1) scanf("%lld",&num[j]);
        printf("Case #%d: ",i);
        sum += n * b + 2 * a;
        if(n >= 2)
        {
            go(j,1,n-1)
            {
                if((num[j] - num[j-1] - 1)*b <= (a*2))
                    sum += (num[j] - num[j-1] - 1)*b;
                else
                    sum += a*2;
            }
            printf("%lld\n",sum);
        }
        else
            printf("%lld\n",sum);
    }
    return 0;
}