zoukankan      html  css  js  c++  java
  • uva 558 tree(不忍吐槽的题目名)——yhx

    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.

     1 #include<cstdio>
     2 #include<cstring>
     3 struct node
     4 {
     5     int lch,rch;
     6 }a[10010];
     7 int zx[10010],hx[10010],n,ans=0x7f7f7f7f,ans_leaf=0x7f7f7f7f;
     8 bool rdzx()
     9 {
    10     int i,j,k,p,q,x,y,z;
    11     char c;
    12     if (scanf("%d",&zx[1])==-1) return 0;
    13     n=1;
    14     while (scanf("%c",&c)&&c==' ')
    15       scanf("%d",&zx[++n]);
    16     return 1;
    17 }
    18 void rdhx()
    19 {
    20     int i;
    21     for (i=1;i<=n;i++)
    22       scanf("%d",&hx[i]);
    23 }
    24 int crt(int lz,int rz,int lh,int rh)
    25 {
    26     if (lz>rz) return 0;    //注意终止条件
    27     int i,j,k,p,q,x,y,z;
    28     p=hx[rh];
    29     i=lz;
    30     while (zx[i]!=p) i++;
    31     a[p].lch=crt(lz,i-1,lh,i+lh-lz-1);   //关键在于这两个式子的推导,列方程求比较不费脑子
    32     a[p].rch=crt(i+1,rz,rh-rz+i,rh-1);
    33     return p;
    34 }
    35 void sc(int p,int x)
    36 {
    37     int i,j,k,l,m,y,z;
    38     x+=p;
    39     if (!a[p].lch&&!a[p].rch)
    40     {
    41         if (x<ans||(x==ans&&p<ans_leaf))
    42         {
    43             ans=x;
    44             ans_leaf=p;
    45             return;
    46         }
    47     }
    48     if (a[p].lch) sc(a[p].lch,x);
    49     if (a[p].rch) sc(a[p].rch,x);
    50 }
    51 int main()
    52 {
    53     int i,j,k,l,m,p,q,x,y,z;
    54     while (rdzx())
    55     {
    56         rdhx();
    57         memset(a,0,sizeof(a));
    58         crt(1,n,1,n);
    59         ans=ans_leaf=0x7f7f7f7f;
    60         sc(hx[n],0);
    61         printf("%d
    ",ans_leaf);
    62     }
    63 }

    递归求树,由于后序遍历的最后一个元素一定是根,所以可以在中序中找到根的位置,则其左边为左子树,右边为右子树,分治求解。

    求值相对简单,dfs即可。

  • 相关阅读:
    关于POST的请求的问题的汇总
    EF关于报错Self referencing loop detected with type的原因以及解决办法
    系统体验设计需注意的地方
    win10文件共享的实现
    idea快速生成实体类Entity
    idea下使用码云插件进行git提交
    漏洞利用-FTP漏洞利用
    metasploit2
    day02-业务服务监控
    day01-系统基础信息模块
  • 原文地址:https://www.cnblogs.com/SBSOI/p/5575050.html
Copyright © 2011-2022 走看看