HDU 1312:Red and Black Time Limit:1000MS     Memory Limit:30000KB     64bit IO Format:%I64d & %I64u   Description There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. F…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1312 Red and Black Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 17773    Accepted Submission(s): 10826 Problem Description There is a rectangula…

Red and Black Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8435    Accepted Submission(s): 5248 Problem Description There is a rectangular room, covered with square tiles. Each tile is colore…

http://poj.org/problem?id=1979 基础搜索. #include <iostream> #include <cstdio> #include <cmath> #include <vector> #include <cstring> #include <string> #include <algorithm> #include <string> #include <set>…

搜索虐我千万遍@_@-----一道搜索的水题,WA了好多好多次@_@发现是n,m搞反了-_- 题意-- 给出m行 n列的矩形,其中从@出发,不能跳到#,只能跳到'.'问最多能够跳到多少块'.' 直接搜就好,不用剪枝 #include <stdio.h> #include <string.h> int n,m,cnt; char map[1000][1000]; int dir[4][2]={{1,0},{0,1},{-1,0},{0,-1}}; void dfs(int x,int…

HDU 1312 题目大意: 一个地图里面有三种元素,分别为"@",".","#",其中@为人的起始位置,"#"可以想象为墙,然后.为可以走的空地,求人可以走的最大点数. 解题思路:从起点开始,起点的四个方向满足条件的点分别入队(放置重复入队,需只要一入队就标记已访问而不是取出时再进行标记),直至队为空. /*HDU 1312 ----- Red and Black 入门搜索 BFS解法*/ #include <cstd…

HDU 1312 题目大意: 一个地图里面有三种元素,分别为"@",".","#",其中@为人的起始位置,"#"可以想象为墙,然后.为可以走的空地,求人可以走的最大点数. 解题思路:从起点开始,从4个方向进行递归遍历(已经访问的点进行标记). /*HDU 1312 ----- Red and Black 入门搜索 */ #include <cstdio> int n, m; //n行m列 int cnt, star…

传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1312 Red and Black Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 25397    Accepted Submission(s): 15306 Problem Description There is a rectangula…

题目代号:HDU 1312 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1312 Red and Black Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 20820    Accepted Submission(s): 12673 Problem Description There i…

嗯... 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1312 一道很经典的dfs,设置上下左右四个方向,读入时记下起点,然后跑dfs即可...最后答案加上起点的1,无需回溯!! AC代码: #include<cstdio> #include<iostream> #include<cstring> using namespace std; ][] = {{-, }, {, }, {, -}, {, }}; int n,…