NOIP 2004 普及组 sdnu 1168.FBI树

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原题链接：

http://210.44.14.31/problem/show/1168

考查树的构造和后序遍历。

代码如下：

#include<iostream>#include<cmath>#include<string>#include<cstring>using namespace std;typedef struct node{char fbi;struct node * leftchild, *rightchild;node() :leftchild(NULL), rightchild(NULL){}//构造函数的简写}*NODE;string numstr;void CreatFbiTree(int left,int right,NODE parent){if (left == right)//递归结束条件{if (numstr[left] == '0')parent->fbi = 'B';else parent->fbi = 'I';return;}bool b = false, i = false;//判断0，1for (int j = left; j <= right; j++){if (numstr[j] == '0'&&!b)b = true;else if (numstr[j] == '1'&&!i)i = true;if (i&&b) break;}if (i&&b)parent->fbi = 'F';else if (!i&&b)parent->fbi = 'B';elseparent->fbi = 'I';if (left <= (right + left) / 2){parent->leftchild = new node();CreatFbiTree(left, (right + left) / 2, parent->leftchild);}if ((right + left) / 2 + 1 <= right){parent->rightchild = new node();CreatFbiTree((right + left) / 2 + 1, right, parent->rightchild);}return;}void PostTraverseTree(NODE root)//后序遍历{if (NULL != root->leftchild)PostTraverseTree(root->leftchild);if (NULL != root->rightchild)PostTraverseTree(root->rightchild);cout << root->fbi;return;}int main(){int n;cin >> n;NODE root=new node();cin >> numstr;CreatFbiTree(0, numstr.length() - 1, root);PostTraverseTree(root);cout << endl;return 0;}

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