Dormitory's Elevator
Accepted : 70	?	Submit : 394
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

解題思路：這個題最開始想開二維dp[i][j]，一看數據就pass了。然后想了下，dp[i]表示電梯停到第i層樓時，學生到1-i層樓的最小化費。這樣為了保證最小花費，肯定要盡可能多的使用電梯，那么dp[i]的狀態首先肯定是可以從dp[i-2]轉移過來的，如果僅僅只有dp[i-2]，那么電梯就可能會停到2-4-6.....，中間停奇數層的可能就被忽略了。所以會有dp[i-3]的情況，保證了奇偶層都可能停，dp[i-4]就沒有必要了，因為肯定會坐電梯上來的。這樣在第i層，i-1層，i-2層之間進行狀態轉移。詳見代碼。 #include<iostream> #include<cstdio> #include<cstring> using namespace std;const int maxn = 100005; int n,m,A,B,dp[maxn],F[maxn],cnt[maxn];int MIN(int a,int b,int c,int d) {if(a <= b && a <= c && a <= d) return a;if(b <= a && b <= c && b <= d) return b;if(c <= a && c <= b && c <= d) return c;if(d <= a && d <= b && d <= c) return d; }int main() {int t,cas = 1;scanf("%d",&t);while(t--){scanf("%d%d%d%d",&n,&m,&A,&B);memset(cnt,0,sizeof(cnt));memset(dp,0,sizeof(dp));for(int i = 1; i <= m; i++){scanf("%d",&F[i]);cnt[F[i]]++;}dp[1] = dp[2] = 0;dp[3] = min(A,B)*cnt[2];for(int i = 4; i <= n; i++){dp[i] = dp[i-2] + min(A,B)*cnt[i-1];int tmp1 = cnt[i-1]*A + cnt[i-2]*2*A; //從i樓下到i-1和i-2樓int tmp2 = cnt[i-1]*2*B + cnt[i-2]*B; //從i-3樓上到i-1和i-2樓int tmp3 = cnt[i-1]*A + cnt[i-2]*B; //i樓下到i-1樓，i-3樓上到i-2樓int tmp4 = cnt[i-2]*2*A + cnt[i-1]*2*B; //i樓下到i-2樓，i-3樓上到i-1樓dp[i] = min(dp[i],dp[i-3] + MIN(tmp1,tmp2,tmp3,tmp4));}int ans = dp[n];ans = min(ans,dp[n-1]+cnt[n]*B);ans = min(ans,dp[n-2]+cnt[n]*2*B + cnt[n-1]*B);printf("Case %d: %d\n",cas++,ans);}return 0; }

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