ACM-ICPC Asia Beijing Regional Contest 2017-J-Pangu and Stones-区间dp-合并石子进阶

ACM-ICPC Asia Beijing Regional Contest 2017-J-Pangu and Stones-区间dp-合并石子进阶

Description

    In Chinese mythology, Pangu is the first living being and the creator of the sky and
 the earth. He woke up from an egg and split the egg into two parts: the sky and the earth.

    At the beginning, there was no mountain on the earth, only stones all over the land.

    There were N piles of stones, numbered from 1 to N. Pangu wanted to merge all of them 
into one pile to build a great mountain. If the sum of stones of some piles was S, Pangu 
would need S seconds to pile them into one pile, and there would be S stones in the new 
pile.

    Unfortunately, every time Pangu could only merge successive piles into one pile. And 
    the number of piles he merged shouldn't be less than L or greater than R.

    Pangu wanted to finish this as soon as possible.

    Can you help him? If there was no solution, you should answer '0'.

Input

    There are multiple test cases.

    The first line of each case contains three integers N,L,R as above mentioned 
    (2<=N<=100,2<=L<=R<=N).

    The second line of each case contains N integers a1,a2 …aN (1<= ai  <=1000,i= 1…N ),
    indicating the number of stones of  pile 1, pile 2 …pile N.

    The number of test cases is less than 110 and there are at most 5 test cases in which N
    >= 50.

Output

For each test case, you should output the minimum time(in seconds) Pangu had to take . If
it was impossible for Pangu to do his job, you should output  0.

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Examples

Input

3 2 2
1 2 3
3 2 3
1 2 3
4 3 3
1 2 3 4

Output

9
6
0

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Problem Description

合并石子进阶版。
给你一堆石子，可以合并连续的X堆，L<=x<=R,每次合并所花费的时间是所合并的石子的数量。求最小花费时间。

Solution

区间dp
先不去管[L,R]的限制，将所有情况通过dp枚举以后，再在题目要求的[L,R]范围里求得答案就好。具体看code。

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Code

/*
 * @Author: Simon 
 * @Date: 2018-08-20 13:30:43 
 * @Last Modified by: Simon
 * @Last Modified time: 2018-08-20 14:43:25
 */
#include
using namespace std;
typedef int Int;
#define int long long
#define INF 0x3f3f3f3f
#define maxn 105
int a[maxn];
int dp[maxn][maxn][maxn];//dp[i][j][k],表示将i~j合并为k堆所需的最小花费
Int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n,l,r;
    while(cin>>n>>l>>r)
    {
        memset(dp,INF,sizeof(dp));
        for(int i=1;i<=n;i++)
        {
            dp[i][i][1]=0;
            cin>>a[i];
            a[i]+=a[i-1];
        }
        for(int x=l-1;x//初始化
        {
            for(int i=1;i+x<=n;i++)
            {
                int j=x+i;
                dp[i][j][x+1]=0;// x堆石子，分为x堆的最小花费为0
                dp[i][j][1]=a[j]-a[i-1];//将所有石子合并为一堆的最小花费为其和
            }
        }
        for(int x=1;x//枚举区间长度
        {
            for(int i=1;i+x<=n;i++)
            {
                int j=x+i;
                for(int k=i;kfor(int c=2;c<=x+1;c++)//先对相应长度进行任意划分
                    {
                        dp[i][j][c]=min(dp[i][j][c],dp[i][k][c-1]+dp[k+1][j][1]);
                    }
                }
                for(int c=l;c<=r;c++)//最后计算满足条件的值
                        dp[i][j][1]=min(dp[i][j][1],dp[i][j][c]+a[j]-a[i-1]);
            }
        }
        if(dp[1][n][1]>=INF)
        {
            cout<<0<else cout<1][n][1]<cin.get(),cin.get();
    return 0;
}