The Settlers of Catan Within Settlers of Catan, the 1995 German game of the year, players attempt to dominate an island by building roads, settlements and cities across its uncharted wilderness. You are employed by a software company that just has de…

Judge Info Memory Limit: 65536KB Case Time Limit: 3000MS Time Limit: 3000MS Judger: Number Only Judger Description Within Settlers of Catan, the 1995 German game of the year, players attempt to dominate an island by building roads, settlements and ci…

The Settlers of Catan Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 1123   Accepted: 732 Description Within Settlers of Catan, the 1995 German game of the year, players attempt to dominate an island by building roads, settlements and c…

题意:画边求最长链,边不能重复数点可以. 很水,用暴力的dfs即可,因为数据不大. 本来以为可以用floyd进行dp的,后来想想好像不能在有回路上的图跑...于是没去做. #include <cstdio> const int maxn = 30; int e[maxn][maxn]; int vis[maxn][maxn]; int n, m, max; void dfs(int x, int d) { if (max < d) max = d; for (int i = 0; i &…

UVa 167 题意:八行八列的棋盘每行每列都要有一个皇后,每个对角线上最多放一个皇后,让你放八个,使摆放位置上的数字加起来最大. 参考:https://blog.csdn.net/xiaoxiede_wo/article/details/79973171 #include <iostream> #include <cstring> #include <iomanip> using namespace std; ][]; int ans; ][]; void dfs(i…

题目 Volume 0. Getting Started 开始10055 - Hashmat the Brave Warrior 10071 - Back to High School Physics 10300 - Ecological Premium 458 - The Decoder 494 - Kindergarten Counting Game 414 - Machined Surfaces 490 - Rotating Sentences 445 - Marvelous Mazes…

ZOJ 2734 Exchange Cards 题目大意: 给定一个值N,以及一堆卡片,每种卡片有一个值value和数量number.求使用任意张卡片组成N的方式. 例如N = 10 ,cards(10,2)(7,2)(5,3)(2,2)(1,5),则10 = 10,10 =7+3+1,10=5+5… 思路分析: 由于之前做过1204,知道这题就是赤裸裸的搜索,直接用dfs暴力即可求得. 可做的优化处理就是——这个过程是贪心Greedy的,先从大到小这样取过来,所以,可以做一步降序排列的预处理.…

杭电ACM分类: 1001 整数求和 水题1002 C语言实验题——两个数比较 水题1003 1.2.3.4.5... 简单题1004 渊子赛马 排序+贪心的方法归并1005 Hero In Maze 广度搜索1006 Redraiment猜想 数论:容斥定理1007 童年生活二三事 递推题1008 University 简单hash1009 目标柏林 简单模拟题1010 Rails 模拟题(堆栈)1011 Box of Bricks 简单题1012 IMMEDIATE DECODABILITY…

各种杂题,水题,模拟,包括简单数论. 1001 A+B 1002 A+B+C 1009 Fat Cat 1010 The Angle 1011 Unix ls 1012 Decoding Task 1019 Grandpa's Other Estate 1034 Simple Arithmetics 1036 Complete the sequence! 1043 Maya Calendar 1054 Game Prediction 1057 Mileage Bank 1067 Rails 10…

7.29   黄昏时刻 (一) 全排列 建模: 给了数字n 代表从1-n 个数全排列 思路: 1. 输入n,如果n值为‘0’,则退出程序 2. vis[i] 保存 是否对第i个数字进行访问 3. dfs 遍历1-n 个数字的全排列,若重复访问,则越过.直到最终访问结束输出访问的的结果.之后回溯删掉对数字的访问. 优化: none 代码: #include <stdio.h> #include <string.h> ] ,n, vis[]; void dfs(int s) { int…