UVA536 Tree Recovery

基礎到不能再基礎的。。數據結構題？（霧）

pre_order的字符串找到樹根，然後在in_order中找到樹根，將樹劃分爲左右子樹。。遞歸求解即可。

AC代碼如下。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

int left[26];
int right[26];
char pre_order[26];
char in_order[26];


int build(int pre_start,int pre_end,int in_start,int in_end)
{
	int root=pre_order[pre_start]-'A';
	if(pre_end-pre_start==1) return pre_order[pre_start]-'A';
	if(pre_end<=pre_start) return -1;
	int mid=find(&in_order[in_start],&in_order[in_end],pre_order[pre_start])-in_order;
	int pos=pre_start;
	for(int i=in_start;i<mid;i++)
        pos=max(pos,(int)(find(&pre_order[pre_start],&pre_order[pre_end],in_order[i])-pre_order));
        left[root]=build(pre_start+1,pos+1,in_start,mid);
        right[root]=build(pos+1,pre_end,mid+1,in_end);
        return root;
}

void print(int root)
{
	if(root==-1)
		return;
	else
	{
		print(left[root]);
		print(right[root]);
		printf("%c", root+'A');
	}
}

int main()
{
	while(scanf("%s",pre_order)!=EOF)
	{
		memset(right,-1,sizeof(right));
		memset(left,-1,sizeof(left));
		scanf("%s", in_order);
		int l=strlen(in_order);
		int root=build(0,l,0,l);
		print(root);
		printf("\n");
	}
	return 0;
}