P1880 [NOI1995]石子合并

　　把环*2转成链就和n个石子放一排的石子合并一样了。

#include<bits/stdc++.h>
using namespace std;
#define  inf 0x3f3f3f3f
#define ls rt<<1
#define rs (rt<<1)+1
#define ll long long
#define fuck(x) cout<<#x<<"     "<<x<<endl;
const int maxn=1e6+10;
int d[4][2]={1,0,-1,0,0,1,0,-1};
int dp1[205][205],dp2[205][205];
int a[205],sum[maxn];
int main(){
    int n;
    cin>>n;
    for(int i=1;i<=n;i++) scanf("%d",&(a[i])),sum[i]=sum[i-1]+a[i];
    for(int i=n+1;i<=2*n;i++) a[i]=a[i-n],sum[i]=sum[i-1]+a[i];
    for(int len=2;len<=n;len++)
        for(int i=1;i<=2*n-len+1;i++){
            int j=i+len-1;
            dp1[i][j]=-inf;
            dp2[i][j]=inf;
            for(int k=i;k<j;k++)
                dp1[i][j]=max(dp1[i][k]+dp1[k+1][j]+sum[j]-sum[i-1],dp1[i][j]),dp2[i][j]=min(dp2[i][k]+dp2[k+1][j]+sum[j]-sum[i-1],dp2[i][j]);
        }
    int maxx=-inf,minn=inf;
    for(int i=1;i<=n+1;i++)
        maxx=max(maxx,dp1[i][i+n-1]),minn=min(minn,dp2[i][i+n-1]);
    cout<<minn<<endl<<maxx<<endl;
    return 0;
}