SDUT-2169-Sequence（DP）

最新推荐文章于 2024-06-26 16:59:45 发布

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本文探讨了如何将一个整数序列分割成指定数量的连续部分，使得总成本最小。通过动态规划方法求解，给出了解题思路和具体实现。

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题目链接：http://www.sdutacm.org/sdutoj/problem.php?action=showproblem&problemid=2169

Problem Description
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.
Input
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.
Output
For each testcase, output one line containing an integer number denoting the minimal cost of splitting the sequence into exactly M succesive parts.
Sample Input
1
5 2
1 3 2 4 5
Sample Output
117

//#include <bits/stdc++.h>
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <algorithm>
using namespace std;
int Scan()
{
    int res=0,ch,flag=0;
    if((ch=getchar())=='-')flag=1;
    else if(ch>='0'&&ch<='9')res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')res=res*10+ch-'0';
    return flag?-res:res;
}
void Out(int a)
{
    if(a>9)Out(a/10);
    putchar(a%10+'0');
}
#define INF 0x3f3f3f3f
#define LL long long
#define bug cout<<"bug"<<endl
const int MAXM = 2e4+7;
const int MAXN = 1007;
long long dp[MAXN][MAXN];
long long sum[MAXN];
int main()
{
    int T,n,m,q,a;
    T=Scan();
    while(T--)
    {
        scanf("%d%d",&n,&m);
        sum[0]=0;
        memset(dp,INF,sizeof(dp));
        for(int i=1; i<=n; ++i)
        {
            scanf("%d",&a);
            sum[i]=sum[i-1]+a;
            dp[1][i]=sum[i]*sum[i];
        }
        for(int i=2; i<=m; ++i)
            for(int j=i; j<=n-m+i; ++j)
                for(int k=i-1; k<=j; ++k)
                    dp[i][j]=min(dp[i][j],dp[i-1][k-1]+(sum[j]-sum[k-1])*(sum[j]-sum[k-1]));
        printf("%lld\n",dp[m][n]);
    }
    return 0;
}

/*
1
5 3
1 3 2 4 5
*/