Working in Beijing

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 659    Accepted Submission(s): 333


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
The 36th ACM/ICPC Asia Regional Shanghai Site —— Warmup
 

Recommend
lcy
 
 
#include<stdio.h>
int main()
{
    long long res;
    int n,a,b;
    int t1,t2;
    int i,T;
    int iCase=0;
    scanf("%d",&T);
    while(T--)
    {
        iCase++;
        scanf("%d%d%d",&n,&a,&b);
        res=a+b;
        scanf("%d",&t1);
        for(i=1;i<n;i++)
        {
            scanf("%d",&t2);
            if((t2-t1-1)*b>2*a)  res+=2*a+b;
            else  res+=b*(t2-t1);
            t1=t2;
        }    
        res+=a;
        printf("Case #%d: %I64d\n",iCase,res);
    }    
    return 0;
}

文章来源:http://www.cnblogs.com/kuangbin/archive/2011/10/04/2198939.html