Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

In a skyscraper people take elevators to go to their destination floors. How-
ever if elevators are not well placed at the beginning, people will spend a
lot of time waiting for the elevators. If an elevator takes a unit of time to
move from one floor to the next floor, what is the minimum possible ex-
pected waiting time for all requests? Formally when we are given the height
of the skyscraper (H), the number of elevators (E), the initial placement of
E elevators, and the pro ability that requests come from each floor, how to
minimize the expected waiting time for all requests?
For example, we assume that there are two elevators in a skyscraper of
eleven floors, and the ratio among the probability that a request comes from
a particular floor is 10 : 1 : 1 : 1 : 1 : 1 : 1 : 1 : 1 : 1 : 1. Now if we place the
elevators on the first and the eighth floors, then in average a request will be
served in 9/10 unit of time. Let pi be the probability that a request comes from
floor i. In this example, p1 = 10/20 = 0.5 and all pi from all other floors will
be 1/20 = 0.05. We also assume that a request will be served by the nearest
elevator, so the request from the fifth to the eleventh floor will be severed by
the elevator at the eight floor, and the requests from the first the to fourth
floor will be severed by the elevator at the first floor. As the result in average
a request will be served in P1i4 |i − 1| × pi + P5i11 |i − 8| × pi = 9/10 .

Technical Specification
Input Format
There are several test cases (no more than 20). The first line of a test case
has two integers, H and E. The next line has H integers for the probability
ratio from the first to the N-th floor.

Output Format
For each test case output the minimum expected waiting time in simple
fraction form. If the denominator is 1 you only output the numerator.

Sample Input
11 2
10 1 1 1 1 1 1 1 1 1 1
11 2
1 0 0 0 0 0 0 0 0 0 1
6 3
0 0 0 0 0 0
3 6
9 9 9

Sample Output
9/10
0
0
0

---

題目意思 : 計算 E 台電梯在 H 樓層, 等待要求最小的期望值
作法 : DP
DP[i][k] //現在有 i 台電梯, 而最後一台電梯的位置 k
我們可以得到
DP[i+1][s] = min(DP[i][k] + 期望值) //s > k
期望值要重新算的部分為
k->(s+k/2) 分給位置在 k 的電梯
(s+k/2)->h 分給位置在 s 的電梯


#include<stdio.h>
long long gcd(long long x, long long y) {
    long long tmp;
    while(x%y) {
        tmp = x, x = y, y = tmp % y;
    }
    return y;
}
long long ER[1001][1001], EL[1001][1001];
int main() {
    int H, E, i, j, k, l, m;
    int Ele[1001] = {0};
    while(scanf("%d %d", &H, &E) == 2) {
        long long sum = 0, cost;
        long long DP[21][1001] = {0}, min;
        for(i = 1; i <= H; i++)
            scanf("%d", &Ele[i]), sum += Ele[i];
        if(sum == 0 || E >= H)    {puts("0");continue;}
        for(i = 1; i <= H; i++) {
            long long tmpsum = 0;
            for(j = i; j <= H; j++) {
                tmpsum += (j-i)*Ele[j];
                ER[i][j] = tmpsum;
            }
        }
        for(i = 1; i <= H; i++) {
            long long tmpsum = 0;
            for(j = i; j >= 1; j--) {
                tmpsum += (i-j)*Ele[j];
                EL[i][j] = tmpsum;
            }
        }
        for(i = 1; i <= E; i++) { /*i台*/
            for(j = 1; j <= H; j++) {/*j位置*/
                min = 0xffffffff;
                for(k = 0; k < j; k++) {/*i-1台 k 位置*/
                    if(i == 1 && k > 0)    break;
                    if(i > 1 && k == 0)    continue;
                    cost = DP[i-1][k];
                    if(i > 1) {
/*                        for(l = k; l <= H; l++)
                            cost -= abs(l-k)*Ele[l];*/
                        cost -= ER[k][H];
                    }
                    m = (k+j)/2;
                    if(i == 1)    m = 0;
                    cost += ER[k][m];
                    cost += EL[j][m+1];
                    cost += ER[j][H];
/*                    for(l = k; l <= m; l++)
                        cost += abs(l-k)*Ele[l];
                    for(l = m+1; l <= H; l++)
                        cost += abs(l-j)*Ele[l];
                        */
                    if(cost < min)    min = cost;
                }
                DP[i][j] = min;
            }
        }
        min = 0xffffffff;
        for(i = 1; i <= H; i++)
            if(min > DP[E][i])
                min = DP[E][i];
        long long    x = min, y = sum, Gcd = gcd(x, y);
        x /= Gcd, y /= Gcd;
        if(y == 1)
            printf("%lld\n", x);
        else
            printf("%lld/%lld\n", x, y);
    }
    return 0;
}