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;}