you_are_the_one(区间dp)

You Are the One

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6711    Accepted Submission(s): 3341

Problem Description

　　The TV shows such as You Are the One has been very popular. In order to meet the need of boys who are still single, TJUT hold the show itself. The show is hold in the Small hall, so it attract a lot of boys and girls. Now there are n boys enrolling in. At the beginning, the n boys stand in a row and go to the stage one by one. However, the director suddenly knows that very boy has a value of diaosi D, if the boy is k-th one go to the stage, the unhappiness of him will be (k-1)*D, because he has to wait for (k-1) people. Luckily, there is a dark room in the Small hall, so the director can put the boy into the dark room temporarily and let the boys behind his go to stage before him. For the dark room is very narrow, the boy who first get into dark room has to leave last. The director wants to change the order of boys by the dark room, so the summary of unhappiness will be least. Can you help him?

 

Input

　　The first line contains a single integer T, the number of test cases.  For each case, the first line is n (0 < n <= 100)
　　The next n line are n integer D1-Dn means the value of diaosi of boys (0 <= Di <= 100)

 

Output

　　For each test case, output the least summary of unhappiness .

 

Sample Input

2 　　 5 1 2 3 4 5 5 5 4 3 2 2

 

Sample Output

Case #1: 20 Case #2: 24

 

Source

2012 ACM/ICPC Asia Regional Tianjin Online

 

Recommend

liuyiding

本题大意：队列中有n个人，要求从1到n依次上台，每个人上台有一个unhappy值val，假设第i个人

第k个上台那么他所带来的unhappy值为(k - 1) * val[ i ]，为了使得unhappy变得足够小，现在允许轮到一个人上台的时候让他进入一个窄巷，窄巷满足先进后出原则，问你通过这个窄巷调整上台顺序，让这n个人都上台后的最小unhappy值为多少。

本题思路：分析容易知道对于第i个人，他是否入栈，何时出栈都会影响到最后的分数，所以我们肯定是要枚举每个人的是否入栈和出入栈状况，我们肯定会枚举每个i，容易想到区间dp，我们用dp[ i ][ j ]表示第 i 到 j 的人已经上台所花费的最小值(仅是i -> j，不考虑其他人)，那么选择断点k，我们让i第k个上台，很容易知道它前面的 k - 1个人已经上台了，i第k各上台的花费为(k - 1) * val[ i ]，他前面的人肯定是已经上台了的，那就是dp[i + 1][i + k - 1]，第k + i ~ j 个人肯定在 i 之后，所以就有了子问题dp[k + i ][ j ]，但是在计算这个子问题时肯定是把i + k当作第1个人的，所以在原问题上我们共把每个数多算了k 次，所以就可以得到状态转移方程dp[ i ][ j ] = (k - 1) * val[ i ] + dp[i + 1][i + k - 1] + dp[i + k][ j ] + (sum[ j ] - sum[i + k - 1]) * k。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int maxn = 200 + 5, inf = 0x3f3f3f3f;

int val[maxn], sum[maxn], dp[maxn][maxn];
int n;

int main() {
    int t, _case = 0;
    scanf("%d", &t);
    while(t --) {
        memset(dp, 0, sizeof dp);
        scanf("%d", &n);
        for(int i = 1; i <= n; i ++) {
            scanf("%d", &val[i]);
            sum[i] = sum[i - 1] + val[i];
        }
        for(int len = 1; len < n; len ++) {
            for(int i = 1; i + len <= n; i ++) {
                int j = i + len;
                dp[i][j] = inf;
                for(int k = 1; k <= len + 1; k ++) {
                    dp[i][j] = min(dp[i][j], (k - 1) * val[i] + dp[i + 1][i + k - 1] + dp[i + k][j] + (sum[j] - sum[i + k - 1] ) * k);
                }
            }
        }
        printf("Case #%d: %d
", ++ _case, dp[1][n]);
    }
    return 0;
}