【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fi,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f1,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;}