題目描述：

思路：
因為n並不大，所以很容易想到列舉一下第一步去掉的邊。
然後展開應為符號和點交雜的一條線，如果符號只有‘+’，那麼用區間dp求其最優解應該很簡單啦，同樣小區間推大區間，dp[i][j] = dp[i][k-1]+dp[k][j].現在考慮‘’，不能向‘+’一樣簡單的相乘就可以，因為如果兩邊是負數，那麼兩邊最小才能使值最大，那麼我們在過程最小值和最大值都需要記憶化，在‘’時，列舉四種:(minn*minn , maxn*minn , maxn*maxn , minn*maxn)取最優即可。

#include <cstdio>
#include <cstring>
 

#include <algorithm>
using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 110;

int minn[N][N],maxn[N][N];
int record[N];
char ver[N];
int a[N];
int main()
{
    int n;
    scanf("%d",&n);
    for(int i = 1; i <= n; i++)
    {
        getchar();
        scanf("%c %d",&ver[i],&a[i]);
        ver[i+n] = ver[i];
        a[i+n] = a[i];
    }
    int
 
 ans = -INF;
    for(int v = 1; v <= n; v++)
    {
        int x = v, y = v+n-1;
        memset(minn,0,sizeof(minn));
        memset(maxn,0,sizeof(maxn));
        for(int i = x; i <= y; i++)
            minn[i][i] = maxn[i][i] = a[i];
        for(int len = 1; len < n; len++)
        {
            for
 
(int i = x; i+len <= y; i++)
            {
                int l = i, r = i+len;
                maxn[l][r] = -INF;
                minn[l][r] = INF;
                for(int k = l+1; k <= r; k++)
                {
                    if(ver[k] == 't')
                    {
                        maxn[l][r] = max(maxn[l][r],maxn[l][k-1]+maxn[k][r]);
                        minn[l][r] = min(minn[l][r],minn[l][k-1]+minn[k][r]);
                    }
                    else
                    {
                        maxn[l][r] = max(maxn[l][r],maxn[l][k-1]*maxn[k][r]);
                        minn[l][r] = min(minn[l][r],maxn[l][k-1]*maxn[k][r]);
                        maxn[l][r] = max(maxn[l][r],maxn[l][k-1]*minn[k][r]);
                        minn[l][r] = min(minn[l][r],maxn[l][k-1]*minn[k][r]);
                        maxn[l][r] = max(maxn[l][r],minn[l][k-1]*minn[k][r]);
                        minn[l][r] = min(minn[l][r],minn[l][k-1]*minn[k][r]);
                        maxn[l][r] = max(maxn[l][r],minn[l][k-1]*maxn[k][r]);
                        minn[l][r] = min(minn[l][r],minn[l][k-1]*maxn[k][r]);
                    }
                }
            }
        }
        //printf("%d\n",maxn[x][y]);
        record[v] = maxn[x][y];
        if(maxn[x][y] > ans)
            ans = maxn[x][y];
    }
    printf("%d\n",ans);
    int flag = 0;
    for(int v = 1; v <= n; v++)
    {
        if(record[v] == ans)
        {
            if(flag == 0)
                printf("%d",v);
            else
                printf(" %d",v);
            flag = 1;
        }
    }
    puts("");
    return 0;
}