[AcWing 1068] 环形石子合并

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点击查看代码

#include<iostream>
#include<cstring>

using namespace std;

const int N = 410;
const int INF = 0x3f3f3f3f;

int n;
int w[N], s[N];
int f[N][N], g[N][N];

int main()
{
    cin >> n;
    for (int i = 1; i <= n; i ++) {
        cin >> w[i];
        w[i + n] = w[i];
    }
    for (int i = 1; i <= 2 * n; i ++)
        s[i] = s[i - 1] + w[i];
    memset(f, 0x3f, sizeof f);
    memset(g, -0x3f, sizeof g);
    for (int len = 1; len <= n; len ++)
        for (int l = 1; l + len - 1 <= 2 * n; l ++) {
            int r = l + len - 1;
            if (len == 1) {
                f[l][r] = g[l][r] = 0;
                continue;
            }
            for (int k = l; k < r; k ++) {
                f[l][r] = min(f[l][r], f[l][k] + f[k + 1][r] + s[r] - s[l - 1]);
                g[l][r] = max(g[l][r], g[l][k] + g[k + 1][r] + s[r] - s[l - 1]);
            }
        }
    int res1 = INF, res2 = -INF;
    for (int i = 1; i <= n; i ++) {
        res1 = min(res1, f[i][i + n - 1]);
        res2 = max(res2, g[i][i + n - 1]);
    }
    cout << res1 << endl;
    cout << res2 << endl;
    return 0;
}

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1. 状态表示
   $f[i][j]$ 表示将第 $i$ 堆石子到第 $j$ 堆石子合并的最小代价
   $g[i][j]$ 表示将第 $i$ 堆石子到第 $j$ 堆石子合并的最大代价
2. 状态计算
   在 $[l,r]$ 区间任找一条分割线 $k$，将区间分为两半
   $f[l][r] = min(f[l][k] + f[k + 1][r] + s[r] - s[l - 1])$
   $g[l][r] = max(g[l][k] + g[k + 1][r] + s[r] - s[l - 1])$
3. 环的处理
   用一个和原数组相同的数组接到数组的末尾，使得数组的长度变为 $2n$，所有环形的方案都可以映射到长度为 $2n$ 的数组上