UVA_548Tree

这是一个很经典的建树,然而当时不会!!!!

给你一个中序和后序 先建一个二叉树,然后找最优解(最优解就是一个叶子节点到根节点权值最小, 同时本身权值最小)

//生成一棵树

int build(int L1, int R1, int L2, int R2) {	//表示1为中序,2为后序 生成这个二叉树  其中R2为这棵树的根
    if(L1 > R1) return 0;
    int root = post_order[R2];
    int p = L1;
    while(in_order[p] != root) ++p;		//在中序中找到根的位置
    int cnt = p - L1;				//cnt表示当前这棵树左子树的大小
    lch[root] = build(L1, p-1, L2, L2 + cnt -1);//这里要知道在后序中,左子树(cnt),右子树,根
    rch[root] = build(p+1, R1, L2+cnt, R2-1);	//那么我们就能知道他们的划分位置,继续递归生成树
    return root;
}

//遍历这棵树

int best,best_sum;

void dfs(int u, int sum) {
    sum += u;
    if(!lch[u] && !rch[u]) {	//叶子节点
        if(sum < best_sum || (sum == best_sum && u < best)){
            best = u;best_sum = sum;
        }
    }
    if(lch[u]) dfs(lch[u], sum);	//接着递归
    if(rch[u]) dfs(rch[u], sum);

}

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<queue>
#include<cstdlib>
#include<algorithm>
#include<stack>
#include<map>
#include<queue>
#include<vector>
#include<sstream>
using namespace std;
const int maxn = 1e4+100;
int in_order[maxn], post_order[maxn], lch[maxn], rch[maxn];
int n;
bool read_list(int* a) {
    string line;
    if(!getline(cin, line)) return false;
    stringstream ss(line);
    n = 0;
    int x;
    while(ss>>x) a[n++] = x;
    return n > 0;
}

int build(int L1, int R1, int L2, int R2) {
    if(L1 > R1) return 0;
    int root = post_order[R2];
    int p = L1;
    while(in_order[p] != root) ++p;
    int cnt = p - L1;
    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;

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(){
#ifdef LOCAL
    freopen("in.txt","r",stdin);
   // freopen("out.txt","w",stdout);
 #endif
    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 << "\n";
    }
    return 0;
}