Description

You are to determine the value of the leaf node in a given binary tree that is the terminal node of a path of least value from the root of the binary tree to any leaf. The value of a path is the sum of values of nodes along that path.

Input

The input file will contain a description of the binary tree given as the inorder and postorder traversal sequences of that tree. Your program will read two line (until end of file) from the input file. The first line will contain the sequence of values associated with an inorder traversal of the tree and the second line will contain the sequence of values associated with a postorder traversal of the tree. All values will be different, greater than zero and less than 10000. You may assume that no binary tree will have more than 10000 nodes or less than 1 node.

Output

For each tree description you should output the value of the leaf node of a path of least value. In the case of multiple paths of least value you should pick the one with the least value on the terminal node.

Sample Input

3 2 1 4 5 7 6
3 1 2 5 6 7 4
7 8 11 3 5 16 12 18
8 3 11 7 16 18 12 5
255
255

Sample Output

1
3
255
 #include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <iomanip>
#include <cstdlib>
#include <sstream>
using namespace std;
const int INF=0x5fffffff;
const double EXP=1e-;
const int mod=;
const int MS=; int in_order[MS],post_order[MS];
int lch[MS],rch[MS];
int n,best,best_sum; bool input()
{
string str;
if(!getline(cin,str))
return false;
stringstream ss(str);
int cnt=;
int x;
while(ss>>x)
in_order[cnt++]=x;
getline(cin,str);
stringstream s(str);
n=cnt;
cnt=;
while(s>>x)
post_order[cnt++]=x;
best_sum=INF;
return true;
} int build(int L1,int R1,int L2,int R2)
{ //返回树根
if(L1>R1)
return ;
int root=post_order[R2];
int p=L1;
while(in_order[p]!=root)
p++;
int cnt=p-L1;
lch[root]=build(L1,p-,L2,L2+cnt-);
rch[root]=build(p+,R1,L2+cnt,R2-);
return root;
} void dfs(int root,int sum)
{
sum+=root;
if(!lch[root]&&!rch[root])
{
if(sum<best_sum||(sum==best_sum&&root<best))
{
best=root;
best_sum=sum;
}
}
if(lch[root])
dfs(lch[root],sum);
if(rch[root])
dfs(rch[root],sum);
} int main()
{
while(input())
{
build(,n-,,n-);
dfs(post_order[n-],);
cout<<best<<endl;
}
return ;
}

F - Tree的更多相关文章

  1. [atcoder contest 010] F - Tree Game

    [atcoder contest 010] F - Tree Game Time limit : 2sec / Memory limit : 256MB Score : 1600 points Pro ...

  2. Codeforces Round #527 (Div. 3) F. Tree with Maximum Cost 【DFS换根 || 树形dp】

    传送门:http://codeforces.com/contest/1092/problem/F F. Tree with Maximum Cost time limit per test 2 sec ...

  3. Codeforces Round #596 (Div. 2, based on Technocup 2020 Elimination Round 2) F. Tree Factory 构造题

    F. Tree Factory Bytelandian Tree Factory produces trees for all kinds of industrial applications. Yo ...

  4. Codeforces Round #499 (Div. 1) F. Tree

    Codeforces Round #499 (Div. 1) F. Tree 题目链接 \(\rm CodeForces\):https://codeforces.com/contest/1010/p ...

  5. AtCoder Grand Contest 010 F - Tree Game

    题目传送门:https://agc010.contest.atcoder.jp/tasks/agc010_f 题目大意: 给定一棵树,每个节点上有\(a_i\)个石子,某个节点上有一个棋子,两人轮流操 ...

  6. Day8 - F - Tree POJ - 1741

    Give a tree with n vertices,each edge has a length(positive integer less than 1001).Define dist(u,v) ...

  7. Codeforces Round #527 (Div. 3) . F Tree with Maximum Cost

    题目链接 题意:给你一棵树,让你找一个顶点iii,使得这个点的∑dis(i,j)∗a[j]\sum dis(i,j)*a[j]∑dis(i,j)∗a[j]最大.dis(i,j)dis(i,j)dis( ...

  8. CF F - Tree with Maximum Cost (树形DP)给出你一颗带点权的树,dist(i, j)的值为节点i到j的距离乘上节点j的权值,让你任意找一个节点v,使得dist(v, i) (1 < i < n)的和最大。输出最大的值。

    题目意思: 给出你一颗带点权的树,dist(i, j)的值为节点i到j的距离乘上节点j的权值,让你任意找一个节点v,使得dist(v, i) (1 < i < n)的和最大.输出最大的值. ...

  9. Codeforces Round #527 F - Tree with Maximum Cost /// 树形DP

    题目大意: 给定一棵树 每个点都有点权 每条边的长度都为1 树上一点到另一点的距离为最短路经过的边的长度总和 树上一点到另一点的花费为距离乘另一点的点权 选定一点出发 使得其他点到该点的花费总和是最大 ...

随机推荐

  1. 【转】MySQL索引和查询优化

    原文链接:http://www.cnblogs.com/mailingfeng/archive/2012/09/26/2704344.html 对于任何DBMS,索引都是进行优化的最主要的因素.对于少 ...

  2. UVALive 5888 Stack Machine Executor (栈+模拟)

    Stack Machine Executor 题目链接: http://acm.hust.edu.cn/vjudge/problem/26636 Description http://7xjob4.c ...

  3. 51Nod 1201 整数划分 (经典dp)

    题目链接:http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1201 题意不多说了. dp[i][j]表示i这个数划分成j个数 ...

  4. zendstudio 出现failed to create the java machine转

    是因为配置java虚拟机内存太小 打开zend for eclipse 10.5时报了个错: failed to create the java virtual machine google了一下,解 ...

  5. cocos2d制作动态光晕效果基础——blendFunc

    转自:http://www.2cto.com/kf/201207/144191.html 最近的项目要求动态光晕的效果. 何谓动态光晕?之前不知道别人怎么称呼这个效果, 不过在我看来,“动态光晕”这个 ...

  6. Vehicle’s communication protocol

    http://www.crecorder.com/techInfo/commuProtocols.jsp COMMUNICATION PROTOCOLS A “communication protoc ...

  7. C166 Interfacing C to Assembler

    Interfacing C to Assembler You can easily interface your C programs to routines written in XC16x/C16 ...

  8. JDBC-ODBC桥乱码问题解决方案

    按照网上提供的ODBC连接数据库的相关资料编写代码,成功编译后运行发现,非中文字段显示正确,而中文字段却是每个汉字以?显示.关于这方面的错误baidu或google下可以找到很多解答方案,我也尝试过其 ...

  9. 用java发送邮件(黄海已测试通过)

    /** * java发送带附件的邮件 * 周枫 * 2013.8.10 */ package com.dsideal.Util; import javax.mail.*; import javax.m ...

  10. iOS 7 Pushing the Limits - Good & Bad Namings in Cocoa

    Cocoa is a dynamically typed language, and you can easily get confused about what type you are worki ...