题目传送门 /* 题意:求从(1, 1)走到(n, m)的二进制路径值最小 BFS+贪心:按照标程的作法,首先BFS搜索所有相邻0的位置,直到1出现.接下去从最靠近终点的1开始, 每一次走一步,不走回头路,只往下或往右走.因为满足i = j + (i - j)的坐标(j, i - j)可能不止一个,选择最小的访问 */ #include <cstdio> #include <algorithm> #include <cstring> #include <cmath…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5335 #include<stdio.h> #include<cstring> #include<cmath> #include<algorithm> using namespace std; ; char R[MAXN][MAXN]; bool used[MAXN][MAXN]; ], yy[]; ] = { , , , -}; ] = { , -, , };…

题目链接: hdu 5335 Walk Out 题目描述: 有一个n*m由0 or 1组成的矩形,探险家要从(1,1)走到(n, m),可以向上下左右四个方向走,但是探险家就是不走寻常路,他想让他所走的路线上的0/1组成的二进数最小,现在要为矫情无比的探险家找最优路径咯. 解题思路: 对于二进制数,前导零是对数字大小没有任何影响的.当到不得不走1的时候就只能向下,或者向右走了.所以先搜索出来一直走零,能走到的最靠近终点的位置,然后在类似搜索,找出最优路径. #include <queue> #…

题目传送门 /* 题意:问一个点到另一个点的最少转向次数. 坐标离散化+BFS:因为数据很大,先对坐标离散化后,三维(有方向的)BFS 关键理解坐标离散化,BFS部分可参考HDOJ_1728 */ #include <cstdio> #include <algorithm> #include <cstring> #include <queue> #include <vector> #include <map> #include <…

Walk Out Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1292    Accepted Submission(s): 239 Problem Description In an n∗m maze, the right-bottom corner is the exit (position (n,m) is the exit).…

Walk Out Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 2912    Accepted Submission(s): 599 Problem Description In an n∗m maze, the right-bottom corner is the exit (position (n,m) is the exit).…

Walk Out                                                                         Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)                                                                                      …

Walk Out Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 141    Accepted Submission(s): 17 Problem Description In an n∗m maze, the right-bottom corner is the exit (position (n,m) is the exit).…

Walk Out Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 639    Accepted Submission(s): 114 Problem Description In an n∗m maze, the right-bottom corner is the exit (position (n,m) is the exit).…

概率DP dp[j][d] 表示不经过i点走d步到j的概率, dp[j][d]=sigma ( dp[k][d-1] * Probability ) ans = sigma ( dp[j][D] ) Walk Time Limit: 30000/15000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 401    Accepted Submission(s): 261 Spe…