IOI1998 Polygon

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Luogu P4342 [IOI1998]Polygon 又是区间dp耶qwq 根据题意，删掉一条边之后合并，其实类似合并果子一题 输入数据，首先任取一点破环称链，对于整个链dp之后，${f_max}[i][i+n-1]$即为答案 本题中，由于涉及乘法运算，可能有两很小的负数相乘，结果却很大，需要再使用$g$数组记录最小值

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#include <bits/stdc++.h>
using namespace std;

const int MAXN = 200;
int f[MAXN][MAXN], g[MAXN][MAXN];
int num[MAXN];
char op[MAXN];

int max(int x, int y, int z, int t, int q)
{
    return max(q, max(max(x, y), max(z, t)));
}

int min(int x, int y, int z, int t, int q)
{
    return min(q, min(min(x, y), min(z, t)));
}

int main()
{
    int n; cin >> n;
    for (int i = 1; i <= n; ++i) {
        cin >> op[i] >> num[i];
        num[i + n] = num[i];
        op[i + n] = op[i];
    }
    memset(f, 0xcf, sizeof(f));
    memset(g, 0x3f, sizeof(g));
    for (int i = 1; i <= n * 2; ++i) {
        f[i][i] = num[i];
        g[i][i] = num[i];
    }
    for (int len = 2; len <= n * 2; ++len) {
        for (int i = 1, j = len; j <= n * 2; ++i, ++j) {
            for (int k = i + 1; k <= j; ++k) {
                if (op[k] == 't') {
                    f[i][j] = max(f[i][j], f[i][k - 1] + f[k][j]);
                    g[i][j] = min(g[i][j], g[i][k - 1] + g[k][j]);
                } else {
                    f[i][j] = max(f[i][j], f[i][k - 1] * f[k][j], 
                        f[i][k - 1] * g[k][j], g[i][k - 1] * f[k][j], g[i][k - 1] * g[k][j]);
                    g[i][j] = min(g[i][j], f[i][k - 1] * f[k][j], 
                        f[i][k - 1] * g[k][j], g[i][k - 1] * f[k][j], g[i][k - 1] * g[k][j]);
                }
            }
        }
    }
    int ans = -(1 << 30);
    for (int i = 1; i <= n; ++i) {
        ans = max(ans, f[i][i + n - 1]);
    }
    cout << ans << endl;
    for (int i = 1; i <= n; ++i) {
        if (f[i][i + n - 1] == ans) {
            cout << i << " ";
        }
    }
    cout << endl;
    return 0;
}

某处g写成f调了半小时…..