XTUOJ 1206 Dormitory's Elevator

Dormitory's Elevator

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

Source

daizhenyang

解题：动态规划，同NYIST的诡异的电梯

#include <bits/stdc++.h>
using namespace std;
const int maxn = 100010;
int dp[maxn],des[maxn];
int main(){
    int T,n,m,A,B;
    scanf("%d",&T);
    for(int t = 1; t <= T; ++t){
        memset(dp,0x3f,sizeof dp);
        memset(des,0,sizeof des);
        scanf("%d %d %d %d",&n,&m,&A,&B);
        for(int i = 0,tmp; i < m; ++i){
            scanf("%d",&tmp);
            des[tmp]++;
        }
        dp[1] = dp[2] = dp[0] = 0;
        for(int i = 3; i <= n; ++i){
            dp[i] = dp[i-2] + min(A,B)*des[i-1];
            int x = min(B,A*2)*des[i-2];
            int y = min(B*2,A)*des[i-1];
            dp[i] = min(dp[i],dp[i-3] + x + y);
        }
        printf("Case %d: %d
",t,dp[n]);
    }
    return 0;
}