xutian_curry的博客

深度优先搜索本题的特点是输入数据直接告诉了图的结点之间是否有边相连，不必要再用邻接矩阵或邻接表去表示#include<stdio.h>#include<stdlib.h>#include <string.h>#include <string>#include <iostream>#include <stack>#i...

深搜 注意搜索规则，注意可以不在当前行放置棋子#include <bits/stdc++.h>using namespace std;int m;//当前已摆好的棋子数int num;//摆放个数int n,k;char Map[15][15];int visited[15];void dfs( int row )//搜索第row行{ if( m == k )...

dfs：每次从元素个数为n的num数组中拿出两个元素做运算，得到的运算结果和剩余n-2元素放入数组进行下一次探索注意 : 浮点数的比较不能直接用==#include <iostream>#include <memory.h>#include <string>#include <stdio.h>#include <queue>#i...

类似城堡问题，注意8个方向都能扩展#include <iostream>#include <memory.h>#include <string>#include <stdio.h>#include <queue>#include <math.h>#include <stdlib.h>using nam

与踩方块完全一样注意回溯#include <iostream>#include <memory.h>#include <string>#include <stdio.h>#include <queue>#include <math.h>#include <stdlib.h>using namespac

求连通子图#include <iostream>#include <memory.h>#include <string>#include <stdio.h>#include <queue>#include <math.h>#include <stdlib.h>using namespace std;

判断从某个点出发是否能走到终点，程序框架#include <iostream>#include <memory.h>#include <string>#include <stdio.h>#include <queue>#include <math.h>#include <stdlib.h>using

#include <iostream>#include <memory.h>#include <string>#include <stdio.h>#include <queue>#include <math.h>#include <stdlib.h>using namespace std;int p

#include<stdio.h>#include<stdlib.h>#include <string.h>#include <string>#include <iostream>#include <stack>#include <math.h>#include <algorithm&a

#include<stdio.h>#include<stdlib.h>#include <string.h>#include <string>#include <iostream>#include <stack>#include <math.h>#include <algorithm&a

如下图所示#include <iostream>#include <vector>#include <stdio.h>#include <memory.h>using namespace std;int visited[9];vector < vector <int> > v(10);void dfs(int i...

#include<stdio.h>#include<stdlib.h>#include <string.h>#include <string>#include <iostream>#include <stack>#include <math.h>#include <algorithm&a

#include <bits/stdc++.h>using namespace std;int a,b;void dfs( int num1, int num2, int step ){ if( num1 % num2 == 0 || num1 / num2 >= 2) { if( step%2 != 1 ) cou...