[NOIP2003]加分二叉树

嘟嘟嘟

中序遍历有一个特点：序列[1, n]是一个中序遍历，i 是根节点，则[1, i - 1]是他的左子树的中序遍历，[i + 1, n]为右子树的中序遍历。所以就想到了区间dp，对于[i, j]枚举根节点k，则dp[i][j] = max(dp[i][k - 1] * dp[k + 1][j] + a[k])。初始化dp[i][i] = a[i]，又因为空节点为1，所以dp[i][i - 1] = 1。

前序遍历的话就再开一个二维数组f[i][j]存这个区间所属哪一个节点，输出的时候跟区间dp很像，具体看代码吧。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define rg register
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 35;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) {last = ch; ch = getchar();}
  while(isdigit(ch)) {ans = ans * 10 + ch - '0'; ch = getchar();}
  if(last == '-') ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}

int n, a[maxn];
ll dp[maxn][maxn];
int f[maxn][maxn];

void print(int L, int R)
{
  if(R < L) return;
  if(L == R) {write(L), space; return;}
  write(f[L][R]), space;
  print(L, f[L][R] - 1);
  print(f[L][R] + 1, R);
}

int main()
{
  n = read();
  for(int i = 1; i <= n; ++i) a[i] = read();
  for(int i = 1; i <= n; ++i) dp[i][i] = a[i], dp[i][i - 1] = 1;
  for(int L = 2; L <= n; ++L)
    for(int i = 1; i + L - 1 <= n; ++i)
      {
    int j = i + L - 1;
    for(int k = i; k <= j; ++k)
      {
        ll Max = dp[i][k - 1] * dp[k + 1][j] + a[k];
        if(Max > dp[i][j]) dp[i][j] = Max, f[i][j] = k;
      }
      }
  write(dp[1][n]); enter;
  print(1, n);
  return 0;
}