You Are the One（区间DP）

You Are the One

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 33 Accepted Submission(s) : 17
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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

#include<iostream>#include<cstring>#include<algorithm>#include<cstdio>#include<string>using namespace std;#define maxsize 105#define INF 0x3f3f3f3fint a[maxsize];int sum[maxsize];int dp[maxsize][maxsize];bool cmp(int a, int b){    return a > b;}int main(){    int t, n, cas = 1;    scanf("%d", &t);    while (t--)    {        scanf("%d", &n);        memset(sum, 0, sizeof(sum));        for (int i = 1; i <= n; i++)        {            scanf("%d", &a[i]);            sum[i] = sum[i - 1] + a[i];        }        memset(dp, 0, sizeof(dp));        for (int i = 1; i <= n; i++)//初始化把每段区间长度设置为无穷大        {            for (int j = i + 1; j <= n; j++)            {                dp[i][j] = INF;            }        }        //n*n*n        for (int p = 1; p<n; p++)//区间长度        {            for (int i = 1; i <= n - 1; i++)//区间起始位置            {                int j = i + p;//末端                for (int k = 1; k <= j - i + 1; k++)//在区间i,j之间，每次决策关一个BOY多久                {                    dp[i][j] = min(dp[i][j], dp[i + 1][i + k - 1] + dp[i + k][j] + k*(sum[j] - sum[i + k - 1]) + a[i] * (k - 1));//对于每一个子区间，可以决策把第一个人放到i+j-1的位置上                }            }        }        printf("Case #%d: %d\n", cas++, dp[1][n]);    }    return 0;}