基本介紹
詳細解釋
圖基本思路
窮舉
系統算法
基本框架
C++的實現
struct Node { int self; //數據 node *left; //左節點 node *right; //右節點 };“ A B C D E F G”
“ const int TREE_SIZE = 9; std::stack<node*> visited, unvisited; node nodes[TREE_SIZE]; node* current; for( int i=0; i<TREE_SIZE; i++) //初始化樹 { nodes[i].self = i; int child = i*2+1; if( child<TREE_SIZE ) //Left child nodes[i].left = &nodes[child]; else nodes[i].left = NULL; child++; if( child<TREE_SIZE ) //Right child nodes[i].right = &nodes[child]; else nodes[i].right = NULL; } unvisited.push(&nodes[0]); //先把0放入UNVISITED stack while(!unvisited.empty()) //只有UNVISITED不空 { current=(unvisited.top()); //當前應該訪問的 unvisited.pop(); if(current->right!=NULL) unvisited.push(current->right); // 把右邊壓入 因為右邊的訪問次序是在左邊之後 if(current->left!=NULL) unvisited.push(current->left); visited.push(current); cout<<current->self<<endl; }”舉例
1 | 1 | 1 | 1 |
0 | 1 | 0 | 1 |
0 | 1 | 0 | 1 |
0 | 1 | 1 | 1 |
constb:array[1..4,1..4]of integer=((1,1,1,1),(0,1,0,1),(0,1,0,1),(0,1,1,1));c:array[1..4,1..2]of -1..1=((0,1),(0,-1),(1,0),(-1,0));vara:array[1..16,1..2]of integer;procedure print;vari,j:integer;beginfor i:=1 to 4 dobeginfor j:=1 to 4 dowrite(b[i,j]:3);writeln;end;writeln('--------------');end;procedure try(k:integer);vari:integer;beginif (a[k,1]=4)and(a[k,2]=4) thenbeginprint;exit;end;for i:=1 to 4 dobegina[k+1,1]:=a[k,1]+c[i,1];a[k+1,2]:=a[k,2]+c[i,2];if (a[k+1,1]>=1) and (a[k+1,1]<=4 )and (a[k+1,2]>=1) and (a[k+1,2]<=4) and(b[a[k+1,1],a[k+1,2]]=1) thenbeginb[a[k+1,1],a[k+1,2]]:=2;try(k+1);b[a[k+1,1],a[k+1,2]]:=1;end;end;end;begina[1,1]:=1;a[1,2]:=1;b[1,1]:=2;try(1);end.
