Sequence【SDUT2169】——区间划分

Sequence
Time Limit: 1000ms Memory limit: 65536K
题目描述
Given an integer number sequence A of length N (1<=N<=1000), we define f(i,j)=(A[i]+A[i+1]+…+A[j])^2 (i<=j). Now you can split the sequence into exactly M (1<=M<= N) succesive parts, and the cost of a part from A[i] to A[j] is f(i,j). The totle cost is the sum of the cost of each part. Please split the sequence with the minimal cost.
输入
At the first of the input comes an integer t indicates the number of cases to follow. Every case starts with a line containing N ans M. The following N lines are A[1], A[2]…A[N], respectively. 0<=A[i]<=100 for every 1<=i<=N.
输出
For each testcase, output one line containing an integer number denoting the minimal cost of splitting the sequence into exactly M succesive parts.
示例输入
1
5 2
1 3 2 4 5
示例输出
117

将区间分为m段，使得m段的平方和最小。
我们定义一个Dp数组Dp[i][j]表示以i结尾，分为j段的最小和，那么对于区间[L+1,i]，Dp[i][j] = min(Dp[i][j],Dp[L][j-1]+$\sum\limits_{k =L+1}^ia[k]$)

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>

using namespace std;

typedef long long LL;

const int Max = 1100;

const int INF = 0x3f3f3f3f;

int a[Max];

LL sum[Max];

LL Dp[Max][Max];

int main()
{
    int T;

    int n,m;

    scanf("%d",&T);

    while(T--)
    {
        scanf("%d %d",&n,&m);

        sum[0] = 0;

        for(int i = 1;i<=n;i++)
        {
            scanf("%d",&a[i]);

            sum[i] = sum[i-1]+a[i];
        }
        for(int i = 1;i<=n;i++)
        {
            Dp[i][1]  =sum[i]*sum[i];

            for(int j = max(m-n+i,2);j<=i;j++)
            {
                Dp[i][j]  = INF;

                if(j>m)//剪枝
                {
                    break;
                }
                for(int k = i-1;k>=j-1;k--)
                {
                    Dp[i][j] = min(Dp[i][j],Dp[k][j-1]+(sum[i]-sum[k])*(sum[i]-sum[k]));

                    if((sum[i]-sum[k])*(sum[i]-sum[k])>=Dp[i][j])//不必要进行的部分
                    {
                        break;
                    }
                }
            }
        }
        printf("%lld\n",Dp[n][m]);
    }
    return 0;
}