【NOIP考前题目回顾】Luogu P1040

思路

第一问看上去像是个树形结构，但是其实我们可以把它直接看成是一个区间DP。令$f_{i, j}$存储区间$[i, j]$内的最大加分，那么三重循环分别枚举左端点、右端点和中间元素，无脑循环出来的$f_{1, n}$即为$[1, n]$区间内的最大加分。在计算过程中，每次记录一下每个节点的位置，方便第二问输出。

代码

#include <cctype>
#include <climits>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <queue>
#include <utility>

int nextInt()
{
    int num = 0;
    char c;
    bool flag = false;
    while ((c = std::getchar()) == ' ' || c == '\r' || c == '\t' || c == '\n');
    if (c == '-')
        flag = true;
    else
        num = c - 48;
    while (std::isdigit(c = std::getchar()))
        num = num * 10 + c - 48;
    return (flag ? -1 : 1) * num;
}

typedef long long LL;

LL n, f[31][31] = { 0 }, i, j, k, x, a[31] = { 0 }, node[31][31] = { 0 };

void LTR(const int l, const int r)
{
    if (l > r)
        return;
    std::cout << node[l][r] << ' ';
    LTR(l, node[l][r] - 1);
    LTR(node[l][r] + 1, r);
}

int main(int argc, char **argv)
{
    int n = nextInt();
    for (int i = 0; i <= n; i++)
        for (int j = 0; j <= n; j++)
            f[i][j] = 1;
    for (int i = 1; i <= n; i++)
    {
        a[i] = nextInt();
        f[i][i] = a[i];
        node[i][i] = i;
    }
    for (int i = n - 1; i >= 1; i--)
        for (int j = i + 1; j <= n; j++)
            for (int k = i; k <= j; k++)
                if (f[i][k - 1] * f[k + 1][j] + a[k] > f[i][j])
                {
                    node[i][j] = k;
                    f[i][j] = f[i][k - 1] * f[k + 1][j] + a[k];
                }
    std::cout << f[1][n] << std::endl;
    LTR(1, n);
#ifdef __EDWARD_EDIT
    std::cin.get();
    std::cin.get();
#endif
    return 0;
}