2017 ICPC亚洲区域赛北京站 JPangu and Stones（区间dp）

Posted 2021-01-20 kannyi

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 …a_N (1<= a_i  <=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.

Sample Input

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

Sample Output

9
6
0

题意：

n个石子堆排成一排，每次可以将连续的[L,R]堆石子合并成一堆，花费为要合并的石子总数。求将所有石子合并成一堆的最小花费，如无法实现则输出0。

思路：

dp[i][j][k]表示区间[i, j]分成k堆的最小代价，转移有

k=1时：

dp[i][j][1]=min(dp[i][j][1],dp[i][j][q]+sum[j]-sum[i-1]) 

k>1时：

dp[i][j][q]=min(dp[i][j][q],dp[i][k][q-1]+dp[k+1][j][1]) 

#include<bits/stdc++.h> 
using namespace std;
#define MAX 105
#define INF 0x3f3f3f3f
int sum[MAX],dp[MAX][MAX][MAX];
int main()
{
    int n,l,r,i,j,k;
    while(scanf("%d%d%d",&n,&l,&r)!=EOF)
    {
        memset(dp,INF,sizeof(dp));
        for(i=1;i<=n;i++)
        {
            scanf("%d",&sum[i]);
            dp[i][i][1]=0;
            sum[i]+=sum[i-1];
        }
        int len;
        for(len=l;len<=r;len++)    //merge长度 len[l,r] 
        {
            for(i=1;i+len-1<=n;i++)//merge范围 [i,i+len-1]
            {
                j=i+len-1;
                dp[i][j][len]=0;
                dp[i][j][1]=sum[j]-sum[i-1];
            }
        }
    
        int q;
        for(len=2;len<=n;len++)    //merge长度 len[2,n] 
        {
            for(i=1;i+len-1<=n;i++)//merge范围 [i,i+len-1]
            {
                j=i+len-1;
                for(k=i;k<j;k++)
                    for(q=2;q<=len;q++)
                        dp[i][j][q]=min(dp[i][j][q],dp[i][k][q-1]+dp[k+1][j][1]);
                for(q=l;q<=r;q++)
                    dp[i][j][1]=min(dp[i][j][1],dp[i][j][q]+sum[j]-sum[i-1]);
            } 
        }
    
        if(dp[1][n][1]<INF)
            printf("%d
",dp[1][n][1]);
        else printf("0
");
    }
    return 0;
}