#include<algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cstdlib>
#include <string>
#include <vector>
#include <cstdio>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
typedef long long ll;
#define rep(i,a,n) for (int i = a; i < n; ++i)
#define per(i,a,n) for (int i = n-1; i >= a; --i)
#define SZ(x) ((int)x.size())
using namespace std;
//head
//因为各个结点的权值各不相同且都是正整数,直接用权值作为结点编号
const int maxn = 10000 + 10;
int in_order[maxn], post_order[maxn], lch[maxn], rch[maxn];
int n;
bool read_list(int *a)
{
string line;
while (!getline(cin, line)) return 0;
stringstream ss(line);
n = 0;
int x;
while (ss >> x)a[n++] = x;
return n > 0;
}
//把in_order[L1..R1]和post_order[L2..R2]建成一颗二叉树,返回数根
int build(int L1, int R1, int L2, int R2)
{
if (L1 > R1) //空树
return 0;
int root = post_order[R2];
int p = L1;
while (in_order[p] != root) p++;
int cnt = p - L1;//左子树的结点个数
lch[root] = build(L1, p - 1, L2, L2 + cnt - 1);
rch[root] = build(p + 1, R1, L2 + cnt, R2 - 1);
return root;
}
int best, best_sum;//目前为止的最优解和对应的权和
void dfs(int u, int sum)
{
sum += u;
if (!lch[u] && !rch[u])
{
if (sum < best_sum || (sum == best_sum&&u < best))
{
best = u;
best_sum = sum;
}
}
if (lch[u])
dfs(lch[u], sum);
if (rch[u])
dfs(rch[u], sum);
}
int main()
{
while (read_list(in_order))
{
read_list(post_order);
build(0, n - 1, 0, n - 1);
best_sum = 1000000000;
dfs(post_order[n - 1], 0);
cout << best << endl;
}
system("pause");
return 0;
}
这个程序中需要特别拿出来分析的是输入函数:
bool read_list(int *a)
{
string line;
while (!getline(cin, line)) return 0;
stringstream ss(line);
n = 0;
int x;
while (ss >> x)a[n++] = x;
return n > 0;
}
和DFS的用法:
void dfs(int u, int sum)
{
sum += u;
if (!lch[u] && !rch[u])
{
if (sum < best_sum || (sum == best_sum&&u < best))
{
best = u;
best_sum = sum;
}
}
if (lch[u])
dfs(lch[u], sum);
if (rch[u])
dfs(rch[u], sum);
}