题解

菜鸡的我一开始甚至在想是不是模拟搜索/px
DP[i][j]表示区间i~j取得的最小答案,那么枚举区间i~j内第i个人上台的次序就可以划分为dp[i+1][k],dp[k+1][j]两个问题

显然i点的ds值是c[i](k-i) 而i+1~j的ds值为(sum[j]-sum[i])(k-i+1)

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#define rep(i,x,y) for(register int i=x;i<=y;++i)
#define repd(i,x,y) for(register int i=x;i>=y;--i)
#define ll long long
using namespace std;
const int N=107;
const int inf=0x3f3f3f3f;
int dp[N][N],c[N],sum[N],n,T;
int main(){
	scanf("%d",&T);
	rep(cnt,1,T){
		scanf("%d",&n);
		rep(i,1,n){scanf("%d",&c[i]);sum[i]=sum[i-1]+c[i];}
		rep(l,2,n)rep(i,1,n-l+1){
			int j=i+l-1;
			dp[i][j]=inf;
			rep(k,i,j)dp[i][j]=min(dp[i][j],dp[i+1][k]+dp[k+1][j]+c[i]*(k-i)+(k-i+1)*(sum[j]-sum[k]));
		}
		printf("Case #%d: %d\n",cnt,dp[1][n]);
	}
	return 0;
}