Problem Description A DFS(digital factorial sum) number is found by summing the factorial of every digit of a positive integer. For example ,consider the positive integer 145 = 1!+4!+5!, so it's a DFS number. Now you should find out all the DFS numbe…

Problem Description A DFS(digital factorial sum) number is found by summing the factorial of every digit of a positive integer. For example ,consider the positive integer 145 = 1!+4!+5!, so it's a DFS number. Now you should find out all the DFS numbe…

Red and Black Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 7337    Accepted Submission(s): 4591 Problem Description There is a rectangular room, covered with square tiles. Each tile is color…

#include<cstdio> #include<cstring> #include<cmath> ][]; #define inf 0xffffff int n,m; int min; ]={,,,-}; ]={,,-,}; ][]; int front,rear; void dfs(int x,int y,int c_step) { if(x==n&&y==m) { if(c_step<min) {min=c_step;return;} }…

#include<cstdio> #include<cstring> ]={,,,-}; ]={,,-,}; ][]; int x1,y1,x2,y2; int step; int n,m,t; void dfs(int x,int y,int c_step) { if(x==x2&&y==y2&&c_step==t) { step=;return; } if((x>x2?x-x2:x2-x)+(y>y2?y-y2:y2-y)+c_ste…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1427 思路分析: 题目要求判断是否存在一种运算组合使得4个数的计算结果为24,因为搜索的层次为3层,不需要选择出最短的路径,采用dfs更有效: 拓展状态时,从当前状态拥有的数中选取两个进行某种运算(因为两个数之间存在大小关系,所以对于除法与减法来说,运算顺序一定, 大的数为被减数或被除数:加法与乘法具有交换律,相对顺序没有影响),如果可以进行运算且运算结果满足题目要求,则该状态可以 拓展,如此拓展状…

Problem Description A DFS(digital factorial sum) number is found by summing the factorial of every digit of a positive integer. For example ,consider the positive integer 145 = 1!+4!+5!, so it's a DFS number. Now you should find out all the DFS numbe…

题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=2212 DFS Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6322    Accepted Submission(s): 3898 Problem Description A DFS(digital factorial sum) numb…

题目传送门 /* 题意:在一个矩阵里放炮台,满足行列最多只有一个炮台,除非有墙(X)相隔,问最多能放多少个炮台 搜索(DFS):数据小,4 * 4可以用DFS,从(0,0)开始出发,往(n-1,n-1)左下角走,x = cnt / n; y = cnt % n; 更新坐标, 直到所有点走完为止,因为从左边走到右边,只要判断当前点左上方是否满足条件就可以了 注意:当前点不能放炮台的情况也要考虑 g[x][y] == 'o'; 的错误半天才检查出来:) */ #include <cstdio> #…

题目传送门 /* 题意:处理完i问题后去处理j问题,要满足a[i][j] <= a[j][k],问最多能有多少问题可以解决 DFS简单题:以每次处理的问题作为过程(即行数),最多能解决n个问题,相同的问题(行数)不再考虑 详细解释:http://blog.csdn.net/libin56842/article/details/41909429 */ #include <cstdio> #include <iostream> #include <cstring> #…