Description
Zuosige always has bad luck. Recently, he is in hospital because of pneumonia. While he is taking his injection, he feels extremely bored. However, clever Zuosige comes up with a new game.
Zuosige knows there is a typical problem called Merging Stones. In the problem, you have N heaps of stones and you are going to merging them into one heap. The only restriction is that you can only merging adjacent heaps and the cost of a merging operation is the total number of stones in the two heaps merged. Finally, you are asked to answer the minimum cost to accomplish the merging.
However, Zuosige think this problem is too simple, so he changes it. In his problem, the cost of a merging is a polynomial function of the total number of stones in those two heaps and you are asked to answer the minimum cost.
Input
The first line contains one integer T, indicating the number of test cases.
In one test case, there are several lines.
In the first line, there are an integer N (1<=N<=1000).
In the second line, there are N integers. The i-th integer si (1<=si<=40) indicating the number of stones in the i-th heap.
In the third line, there are an integer m (1<=m<=4).
In the forth line, there are m+1 integers a0, … , am. The polynomial function is P(x)= (a0+a1*x+a2*x2+…+am*xm). (1<=ai<=5)
Output
For each test case, output an integer indicating the answer.
Sample Input
1
5
3 1 8 9 9
2
2 1 2
Sample Output
2840
HINT
Source
题意：有n堆石头，给出每堆石头的价值，然后给出m+1个系数，合并两堆石子需要花费(a
0+a
1*x+a
2*x
2+…+a
m*x
m)的能量，问合并所有石子所花费的最小能量
思路：状态压缩，但是我们还需要优化，由于合并必须是相邻的，我们可以再求区间最小值的时候，同时记录合并点，然后我们枚举区间的时候，在两个合并点之间枚举，其中原因大家自己可以理解一下
include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <bitset>
#include <algorithm>
using namespace std;
#define ls 2*i
#define rs 2*i+1
#define up(i,x,y) for(i=x;i<=y;i++)
#define down(i,x,y) for(i=x;i>=y;i--)
#define mem(a,x) memset(a,x,sizeof(a))
#define w(a) while(a)
#define LL long long
const double pi = acos(-1.0);
#define N 1005
#define mod 19999997
#define INF 0x3f3f3f3f
#define exp 1e-8
LL dp[N][N],vec[50000],sum[N];
int s[N],a[N],t,n,m,tot,vis[N][N];
LL col(LL x)
{
LL ans = a[0];
int i,j;
up(i,1,m)
{
LL tem = 1;
up(j,1,i) tem*=x;
ans+=tem*a[i];
}
return ans;
}
int main()
{
int i,j,k;
scanf("%d",&t);
w(t--)
{
scanf("%d",&n);
mem(sum,0);
mem(dp,0);
tot=0;
up(i,1,n)
{
scanf("%d",&s[i]);
sum[i] = sum[i-1]+s[i];
tot+=s[i];
}
scanf("%d",&m);
up(i,0,m)
{
scanf("%d",&a[i]);
}
up(i,1,tot)
{
vec[i]=col((LL)i);
}
up(i,1,n) vis[i][i] = i;
vis[0][1] = 1;
int len;
up(len,2,n)
{
up(i,1,n-len+1)
{
j = i+len-1;
dp[i][j] = 1LL<<60;
up(k,vis[i][j-1],vis[i+1][j])
{
LL tem = dp[i][k]+dp[k+1][j]+vec[sum[j]-sum[i-1]];
if(tem<dp[i][j])
{
dp[i][j] = tem;
vis[i][j] = k; //记录合并点
}
}
}
}
printf("%lld\n",dp[1][n]);
}
return 0;
}
