XTU 1206 Dormitory's Elevator

Dormitory's Elevator
Accepted : 52		Submit : 332
Time Limit : 1000 MS		Memory Limit : 65536 KB

Problem Description The new dormitory has N(1≤N≤100000) floors and M(1≤M≤100000)students. In the new dormitory, in order to save student's time as well as encourage student exercise, the elevator in dormitory will not stop in adjacent floor. So if there are people want to get off the elevator in adjacent floor, one of them must walk one stair instead. Suppose a people go down 1 floor costs A energy, go up 1 floor costs B energy(1≤A,B≤100). Please arrange where the elevator stop to minimize the total cost of student's walking cost.All students and elevator are at floor 1 initially, and the elevator can not godown and can stop at floor 2. Input First line contain an integer T, there are T(1≤T≤10) cases. For each case T, there are two lines. First line: The number of floors N(1≤N≤100000), and the number of students M(1≤M≤100000),A,B(1≤A,B≤100) Second line: M integers (2≤A[i]≤N), the student's desire floor. Output Output case number first, then the answer, the minimum of the total cost of student's walking cost. Sample Input 1 3 2 1 1 2 3 Sample Output Case 1: 1

#include <iostream>
#include <stdio.h>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#define N 100009
#define ll __int64
using namespace std;

ll dp[N];//表示从第一层到第n层的最小体力耗费
int a[N];
int n,m,A,B;

//想要少走，那么就要尽可能多的让电梯停下来
//如果现在在第i层，那么上一次停靠可以是第i-2,或者第i-3层（因为要多停）

int main()
{

    int t;
    scanf("%d",&t);

    for(int ca=1;ca<=t;ca++)
    {
        scanf("%d %d %d %d",&n,&m,&A,&B);

        int ff;
        memset(a,0,sizeof a);
        memset(dp,0,sizeof dp);

        for(int i=0;i<m;i++)
        {
            scanf("%d",&ff);
            a[ff]++;
        }
        dp[1]=0;
        dp[2]=0;//可以停在二楼

        //dp[2]=min(A,B)*a[2];
        //dp[3]=a[2]*A;

        for(int i=3;i<=n;i++)
        {
            ll tt=1<<30;

            ll ff=dp[i-2]+min(A,B)*a[i-1];//当上一次是i-2停时，要去i-1的人，可以选择i下或者i-2的时候下，选小的。
            if(i-3>0)
             tt=dp[i-3]+min(2*B*a[i-1]+B*a[i-2],min( A*a[i-1]+A*a[i-2]*2, min(A*a[i-1]+B*a[i-2],A*a[i-2]*2+B*a[i-1]*2 ) ));
//当上一次是i-3层停时，那么会有下面四种选择
//(i-3->i-1,i-3->i-2); (i->i-1,i->i-2); (i->i-1,i-3->i-2); (i->i-2,i-3->i-1)
            dp[i]=min(ff,tt);
        }

        printf("Case %d: %I64d\n",ca,dp[n]);

    }

    return 0;
}

/*
3
4 8 1 2
1 1 2 2 3 3 4 4
*/
//2