Tree UVA - 548

https://vjudge.net/problem/UVA-548
第一行为中序遍历，第二行为后序遍历
考察这方面的知识 同时加强树构建遍历的能力

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
①结点查找依据：树的后序遍历 最后一个为根结点。
②建树方法：由lch[] rch[]构建关系，类似链表
③找出所有结点：树的性质对每个子树同样有效 所以build函数中可以递归找出每个结点

#include <iostream>#include <cstdio>#include <sstream>#include <set>#include <bitset>  #include <queue>  #include <deque>#include <stack> #include <list>#include <vector>#include <map>#include <string>#include <cstring>#include <cmath>#include <algorithm>using namespace std;typedef long long ll;typedef set<int> Set;typedef vector<int> Vec;typedef set<int>::iterator It;#define mem(s,t) (memset(s,t,sizeof(s)))#define SZ(v) (int(v.size()))#define D(v) (cout<<#v<<" "<<v<<endl)#define MAXN 10010int in_order[MAXN],post_order[MAXN],lch[MAXN],rch[MAXN];int n;int read_list(int *a){    n=0;    int x;    string s;    if(!getline(cin,s)) return 0;    stringstream ss(s);    while(ss>>x)    a[n++]=x;    return n;}    int fi(int root,int s1){//找出root在中序遍历中的位置pos     int pos=s1;    while(in_order[pos]!=root) pos++;    return pos;}int build(int s1,int e1,int s2,int e2){     if(s1>e1)   return 0;    int root=post_order[e2];    int pos = fi(root,s1);    int cnt=pos-s1;    //由此再分出左右子树    lch[root]=build(s1,pos-1,s2,s2+cnt-1);    rch[root]=build(pos+1,e1,s2+cnt,e2-1);    return root;}int best_num=0,best_sum;int dfs(int root,int sum){    sum+=root;    //更新best_sum 和 best_num    if(!lch[root] && !rch[root]){        if(sum<best_sum || (root<best_num && sum == best_sum)){            best_sum=sum;            best_num=root;        }    }    if(lch[root]) dfs(lch[root],sum);    if(rch[root]) dfs(rch[root],sum);}int main(int argc, char *argv[]){    while(read_list(in_order)){        read_list(post_order);        build(0,n-1,0,n-1);        best_sum=1<<30;//注意初始化的位置        dfs(post_order[n-1],0);        cout<<best_num<<endl;    }    return 0;}