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Description
Little Valentine liked playing with binary trees very much. Her favorite game was constructing randomly looking binary trees with capital letters in the nodes.
This is an example of one of her creations:
D
/ / B E
/ \ / \ \
A C G
/
/
F
To record her trees for future generations, she wrote down two strings for each tree: a preorder traversal (root, left subtree, right subtree) and an inorder traversal (left subtree, root, right subtree).
For the tree drawn above the preorder traversal is DBACEGF and the inorder traversal is ABCDEFG.
She thought that such a pair of strings would give enough information to reconstruct the tree later (but she never tried it).
Now, years later, looking again at the strings, she realized that reconstructing the trees was indeed possible, but only because she never had used the same letter twice in the same tree.
However, doing the reconstruction by hand, soon turned out to be tedious.
So now she asks you to write a program that does the job for her!
Input is terminated by end of file.
DBACEGF ABCDEFG BCAD CBAD
ACBFGED CDAB
程序分析：骑士走日，和中国象棋中的马一样走日。（使用广搜BFS）
程序代码：
#include <iostream> #include <queue> #include<string.h> using namespace std; int s1,s2,e1,e2; int tu[8][8]; int xx[8] = {1, 2, 1, 2, -1, -2, -1, -2};//x坐标变化 int yy[8] = {2, 1, -2, -1, 2, 1, -2, -1};//y坐标变化 int bfs() { memset(tu,0,sizeof(tu));//数组值全为0 queue<int> q; int x1,y1,x2,y2; tu[s1][s2]=0;// q.push(s1); q.push(s2); while(!q.empty()) { x1=q.front(); q.pop(); y1=q.front(); q.pop(); if(x1==e1&&y1==e2)//在原地不动 return tu[x1][y1]; for(int i=0; i<8; i++) { x2=x1+xx[i]; y2=y1+yy[i]; if(x2<0||x2>7||y2<0||y2>7||tu[x2][y2]>0) continue; tu[x2][y2]=tu[x1][y1]+1; q.push(x2); q.push(y2); } } return 0; } int main() { char a[3],b[3]; int total; while(cin>>a>>b) { s1=a[0]-‘a‘;//输入的字母，是列，从第一列a开始，减去‘a‘，转换 s2=a[1]-‘1‘;//输入的数字，是行，从的第一行开始，减去‘1‘，转换 e1=b[0]-‘a‘; e2=b[1]-‘1‘; total=bfs(); cout<<"To get from "<<a<<" to "<<b<<" takes "<<total<<" knight moves."<<endl; } return 0; }
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原文地址：http://www.cnblogs.com/yilihua/p/4674699.html
