
DFS
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hdu 1181变形课 java实现
第一次用java写深搜, 菜鸟级别, 求大神指点import java.util.Scanner;public class Main { private static String[] sa = new String[10000]; private static int length; //输入的字符串的个数 private static boolean[] visited; //标志原创 2013-12-20 17:46:29 · 718 阅读 · 0 评论 -
poj2965
The Pilots Brothers' refrigeratorTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 20144 Accepted: 7744 Special JudgeDescriptionThe game “The Pilo原创 2015-05-14 23:51:13 · 433 阅读 · 0 评论 -
poj1753(暴搜+位运算)
Flip GameTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 33180 Accepted: 14513DescriptionFlip game is played on a rectangular 4x4 field with two-sided p原创 2015-05-12 23:31:07 · 404 阅读 · 0 评论 -
hdu2660(dfs)
Accepted NecklaceTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2536 Accepted Submission(s): 1003Problem DescriptionI have N原创 2014-07-16 16:04:59 · 508 阅读 · 0 评论 -
hdu1312
Red and BlackTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8569 Accepted Submission(s): 5327Problem DescriptionThere is a r原创 2014-05-09 22:34:03 · 403 阅读 · 0 评论 -
timus1005
http://acm.timus.ru/problem.aspx?space=1&num=1005#include#include#includeusing namespace std;int n;int visit[20];int a[20];int sum = 0;int min1 = 1<<30;void dfs(int values, int cur){原创 2014-04-24 23:32:09 · 585 阅读 · 0 评论 -
uva216
回溯法#include#include#include#includeusing namespace std;int n;bool visit[150];int a[150];int result[150];double min1;int case1 = 0;struct point{ int x, y;};double sum(int a, int b){原创 2014-04-20 13:11:15 · 499 阅读 · 0 评论 -
uva10160(dfs+状态压缩)
Problem D: Servicing stationsA company offers personal computers for sale in N towns (3 <= N <= 35). The towns are denoted by 1, 2, ..., N. There are direct routes connecting M pairs from among th原创 2014-05-04 00:36:39 · 499 阅读 · 0 评论 -
hdu1427
速算24点Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2868 Accepted Submission(s): 685Problem Description速算24点相信绝大多数人都玩过。就是随机给转载 2014-05-02 19:50:51 · 583 阅读 · 0 评论 -
hdu1175
连连看Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 16678 Accepted Submission(s): 4307Problem Description“连连看”相信很多人都玩过。没玩过也没原创 2014-05-01 23:15:54 · 712 阅读 · 0 评论 -
我初步了解的dfs
dfs 关键点有四个:位置,属性, 传递ts原创 2014-05-02 13:46:18 · 600 阅读 · 0 评论 -
BNU4071
我们知道,一张地图使用不同的颜色标注相邻的省份的(废话……不然就乱了……)。但是我们想知道用n种颜色对m个省填色共有多少种方法。 当然,我们会给出这m个省得相邻情况。Input第一行为两个整数n m表示颜色与省份数量。(n,mOutput输出一行整数,表示可能的填充方法数。Sample Input4 441 21 42 33原创 2014-05-01 20:22:56 · 474 阅读 · 0 评论 -
uva10344(全排列+dfs)
Problem I23 Out of 5Input: standard inputOutput: standard outputTime Limit: 1 secondMemory Limit: 32 MBYour task is to write a program that can decide whether you can find an arithmetic ex原创 2014-05-01 17:21:15 · 473 阅读 · 0 评论 -
蓝桥杯 地宫取宝
标题:矩阵翻硬币 小明先把硬币摆成了一个 n 行 m 列的矩阵。 随后,小明对每一个硬币分别进行一次 Q 操作。 对第x行第y列的硬币进行 Q 操作的定义:将所有第 i*x 行,第 j*y 列的硬币进行翻转。 其中i和j为任意使操作可行的正整数,行号和列号都是从1开始。 当小明对所有硬币都进行了一次 Q 操作原创 2014-04-15 17:47:37 · 2130 阅读 · 1 评论