SDUT 2169 Sequence dp减枝区间划分

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<span style="font-family: Arial, Helvetica, sans-serif;">#include<bits/stdc++.h></span>

#define LL long long#define INF 0x3f3f3f3fusing namespace std;LL dp[1010][1010];LL sum[1010];int main(){    int T;    while(~scanf("%d",&T))    {        while(T--)        {            int n,m;            scanf("%d%d",&n,&m);            int a;            memset(dp,INF,sizeof(dp));            dp[0][0]=0;            for(int i=0;i<=m;i++)                dp[i][0]=0;            for(int i=0;i<=n;i++)                dp[0][i]=0;            for(int i=1;i<=n;i++)            {                scanf("%d",&a);                sum[i]=sum[i-1]+a;                dp[1][i]=sum[i]*sum[i];  /// i个数分成一组            }            for(int i=2;i<=m;i++)            {                ///减枝     ///j<=n-m+i：因为当n个数分成m组时是用不到n个数分成m-1组的情况，                                          ///   最多会用到 小问题n-1个数分成m-1组，同理；所以用n个数算m-1,m-2,m-3,m-4...组都是多余的                for(int j=i;j<=n-m+i;j++)                    {                    for(int k=i-1;k<=j-1;k++)                    {                        dp[i][j]=min(dp[i][j],dp[i-1][k]+(sum[j]-sum[k])*(sum[j]-sum[k]));                    }                }            }            printf("%lld\n",dp[m][n]);            ///可以写成一维                        ///            ///            for(int i=2;i<=m;i++)   ///更快///            {               ///                                      ///                for(int j=i;j<=n-m+i;j++)    ///               {///                    for(int k=j-1;k>=i-1;k--)     ///倒着///                    {///                           if((sum[j]-sum[k])*(sum[j]-sum[k])>=dp[i][j]) break;         ///双重减枝，说明已经找到小的了。

///                        dp[i][j]=min(dp[i][j],dp[i-1][k]+(sum[j]-sum[k])*(sum[j]-sum[k]));///                    }///                }///            }///            printf("%lld\n",dp[m][n]);        }    }    return 0;}

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