例题6-8 树（Tree,UVa 548）（刘汝佳算法竞赛）

#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);
}