UVA.548 Tree(二叉树 DFS)
题意分析
给出一棵树的中序遍历和后序遍历,从所有叶子节点中找到一个使得其到根节点的权值最小。若有多个,输出叶子节点本身权值小的那个节点。
先递归建树,然后DFS求解。
代码总览
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <sstream>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <vector>
#define nmax 100000
#define INF 0x3f3f3f3f
using namespace std;
int inorder[nmax],postorder[nmax];
int lnode[nmax],rnode[nmax];
int i;
bool read_input()
{
memset(inorder,0,sizeof(inorder));
memset(postorder,0,sizeof(postorder));
string s;
if(!getline(cin,s)) return false;
else{
stringstream ss(s);
i = 0;
int x;
while(ss >> x) inorder[i++] = x;
getline(cin,s);
stringstream sss(s);
i = 0;
while(sss>>x) postorder[i++] = x;
}
return true;
}
// 递归建树
int buildtree(int l1,int r1, int l2, int r2)
{
if(l1>r1) return 0;
int root = postorder[r2];
int p = l1;
while(inorder[p] != root) p++;
int cnt = p-l1;
lnode[root] = buildtree(l1,p-1,l2,l2+cnt-1);
rnode[root] = buildtree(p+1,r1,l2+cnt,r2-1);//
return root;
}
int bestsum,best;
void dfs(int u, int sum)
{
sum+=u;
if(!lnode[u] && !rnode[u])
if(sum<bestsum || (sum == bestsum && u<best)){
best = u;
bestsum = sum;
}
if(lnode[u]) dfs(lnode[u],sum);
if(rnode[u]) dfs(rnode[u],sum);
}
int main()
{
//freopen("in.txt","r",stdin);
while(read_input()){
bestsum = INF;best = INF;
buildtree(0,i-1,0,i-1);
dfs(postorder[i-1],0);
cout<<best<<"
";
}
return 0;
}