Red and Black Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 26058   Accepted: 14139 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. From a…

-->Red and Black Descriptions: 有个铺满方形瓷砖的矩形房间,每块瓷砖的颜色非红即黑.某人在一块砖上,他可以移动到相邻的四块砖上.但他只能走黑砖,不能走红砖. 敲个程序统计一下这样可以走到几块红砖上. Input  多组测试用例.每组数组开头有两个正整数W和H:W与H分别表示 x- 与 y- 方向上瓷砖的数量.W和W均不超过20. 还有H行数据,每行包含W个字符.每个字符表示各色瓷砖如下. '.' - 一块黑砖 '#' - 一块红砖 '@' - 一个黑砖上的人(一组数…

POJ 1979 Red and Black (红与黑) Time Limit: 1000MS    Memory Limit: 30000K Description 题目描述 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. From a tile, he can move to…

原文来自于:http://www.infoq.com/cn/news/2014/02/wildfly8-launch Red Hat的JBoss部门今天宣布WildFly 8正式发布.其前身是JBoss Application Server.本次发布的版本完全支持Java EE 7规范,支持Web和Full profile.WildFly同时包含全新的Web服务器Undertow.新的安全特性和补丁系统.补丁系统用于对正在运行的系统进行升级. Undertow是一个Servlet 3.1容器,也…

http://poj.org/problem?id=3321 http://acm.hdu.edu.cn/showproblem.php?pid=3887 POJ 3321: 题意:给出一棵根节点为1的边不一定的树,然后给出问题:询问区间和 或者 节点值更新. HDU 3887: 题意:和POJ 3321的题意差不多,只不过对每个节点询问不包含该节点的区间和 思路:今天才学了下才知道有DFS序这种东西,加上树状数组处理一下区间和 和 节点更新. DFS序大概就是我们在DFS遍历一棵树的时候,在进…

POJ 3252:Round Numbers Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 10099 Accepted: 3669 Description The cows, as you know, have no fingers or thumbs and thus are unable to play Scissors, Paper, Stone' (also known as 'Rock, Paper, Sciss…

1.链接地址: http://bailian.openjudge.cn/practice/1979 http://poj.org/problem?id=1979 2.题目: 总时间限制: 1000ms 内存限制: 65536kB 描述 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.…

题目链接:http://poj.org/problem?id=1979 思路分析:使用DFS解决,与迷宫问题相似:迷宫由于搜索方向只往左或右一个方向,往上或下一个方向,不会出现重复搜索: 在该问题中往四个方向搜索,会重复搜索,所以使用vis表来标记访问过的点,避免重复搜索. 代码如下: #include <iostream> using namespace std; ; int vis[MAX_N][MAX_N]; char map[MAX_N][MAX_N]; int red_count,…

传送门: poj:http://poj.org/problem?id=1979 zoj:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1165 题目大意: 给你初始坐标,标记为'#'的格子不能走,求你能走的所有格子的个数(能走的为'.',初始坐标用'@'表示) 思路: 一看直接DFS就好了嘛.... 好几天没刷题了,回到家来水一发先~ #include<cstdio> #include<cstring> con…

地址 http://poj.org/problem?id=1979 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. From a tile, he can move to one of four adjacent tiles. But he can't mo…

http://poj.org/problem?id=1979 #include <cstdio> #include <cstring> using namespace std; const int maxn = 21; bool vis[maxn][maxn]; char maz[maxn][maxn]; int n,m; const int dx[4] = {1,-1,0,0}; const int dy[4] = {0,0,1,-1}; int ans; bool in(int…

Red and Black Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 50913   Accepted: 27001 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. From a…

红与黑 题目大意:一个人在一个矩形的房子里,可以走黑色区域,不可以走红色区域,从某一个点出发,他最多能走到多少个房间? 不多说,DFS深搜即可,水题 注意一下不要把行和列搞错就好了,我就是那样弄错过一次哈哈哈哈 #include <stdio.h> #include <stdlib.h> #define MAX_N 20 static int dp[MAX_N][MAX_N]; static int startx; static int starty; static int ans…

Problem 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. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can mo…

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. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can move only…

Red and Black Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 27891   Accepted: 15142 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. From a…

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. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can move only…

题目: 简单dfs,没什么好说的 代码: #include <iostream> using namespace std; typedef long long ll; #define INF 2147483647 int w,h; ][]; ][] = {-,,,,,-,,}; ; void dfs(int x,int y){ || x >= h || y < || y >= w || a[x][y] == '#') return; ans++; a[x][y] = '#';…

#include<iostream> #include<cstdio> #include<queue> #include<algorithm> #include<cstring> using namespace std; #define MAXN 21 struct node { int x; int y; }; int n,m,g[MAXN][MAXN]; bool vis[MAXN][MAXN]; ]={,-,,}; ]={,,,-}; in…

标准DFS,统计遍历过程中遇到的黑点个数 #include<cstdio> #include<vector> #include<queue> #include<string> #include<map> #include<iostream> #include<cstring> #include<algorithm> using namespace std; typedef long long LL; const…

一开始理解错题意了,以为是走过的砖不能再重复走,最多能走多少个黑砖,结果写的递归陷入死循环...后来才明白原来可以重复走,问可以到达的磁砖数. #include <iostream> #include <string.h> #include <stdio.h> #include <math.h> #include <algorithm> using namespace std; int w,h,num,ans; ][];//存储地图的信息,为0代…

链接 : Here! 思路 : 简单的搜索, 直接广搜就ok了. /************************************************************************* > File Name: E.cpp > Author: > Mail: > Created Time: 2017年11月26日 星期日 10时51分05秒 ********************************************************…

题目地址: http://poj.org/problem?id=1979  或者  https://vjudge.net/problem/OpenJ_Bailian-2816 Red and Black Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 46793   Accepted: 25201 Description There is a rectangular room, covered with square ti…

题目链接:http://poj.org/problem?id=1979 #include<cstring> #include<iostream> using namespace std; ,h=,sum=; ][]; void DFS(int p,int q) { &&p<h&&q>=&&q<n) { sum++; aa[p][q]='#'; } else return ; DFS(p-,q); DFS(p+,q);…

The Same Game Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5168   Accepted: 1944 Description The game named "Same" is a single person game played on a 10 \Theta 15 board. Each square contains a ball colored red (R), green (G), or…

http://poj.org/problem?id=1459 题意:有np个发电站,nc个消费者,m条边,边有容量限制,发电站有产能上限,消费者有需求上限问最大流量. 思路:S和发电站相连,边权是产能上限,消费者和T相连,边权是需求上限,边的话就按题意加就好了.难点更觉得在于输入..加个空格..边数组要*2,因为有反向边. #include <cstdio> #include <algorithm> #include <iostream> #include <cs…

http://poj.org/problem?id=3436 题意:题意很难懂.给出P N.接下来N行代表N个机器,每一行有2*P+1个数字 第一个数代表容量,第2~P+1个数代表输入,第P+2到2*P+1是代表输出 输入有三种情况,0,1,2.输出有0,1两种情况输入0代表不能有这个接口,1代表必须要有这个接口,2代表这个接口可有可无.输出0代表有这个接口,1代表没有这个接口.大概输出就是像插头,输入像插座,只有接口吻合才可以相连. 思路:比较简单的最大流,主要是理解题意很难,把每台机器拆成输…

http://poj.org/problem?id=2195 题意:有一个地图里面有N个人和N个家,每走一格的花费是1,问让这N个人分别到这N个家的最小花费是多少. 思路:通过这个题目学了最小费用最大流.最小费用最大流是保证在流量最大的情况下,使得费用最小. 建图是把S->人->家->T这些边弄上形成一个网络,边的容量是1(因为一个人只能和一个家匹配),边的费用是曼哈顿距离,反向边的费用是-cost. 算法的思想大概是通过SPFA找增广路径,并且找的时候费用是可以松弛的.当找到这样一条增…

http://poj.org/problem?id=3281 题意:有n头牛,f种食物,d种饮料,每头牛有fnum种喜欢的食物,dnum种喜欢的饮料,每种食物如果给一头牛吃了,那么另一个牛就不能吃这种食物了,饮料也同理,问最多有多少头牛可以吃到它喜欢的饮料和食物. 思路:一开始还以为二分匹配可以做,当然如果只有食物或者饮料其中一种就可以做.难点在于建图.看了下书,因为要保证经过牛的流量是1(每种食物对应分配给一头牛,每种饮料对应分配给一头牛,避免一头牛吃多份),所以要把牛拆成两个点.形成这样的路…

http://poj.org/problem?id=3580 题意:有6种操作,其中有两种之前没做过,就是Revolve操作和Min操作.Revolve一开始想着一个一个删一个一个插,觉得太暴力了,后来发现可以把要放到前面的一段切开,丢到前面去,就和上一题的Cut是一样的了.还有Min操作,一开始特别ZZ地想着只要找keytree的最左边就好了,然后发现并不是那样的,要维护一个 mi 值,一开始两个节点设成 INF,然后 pushup 的时候先把 val 赋给 mi,然后再和左右儿子对比.WA了…