POJ 3050 题意: 1.5*5的方阵中,随意挑一格,记住这个格子的数字 2.可以上下左右走,走5次,每走一次记录下所走格子的数字 3.经过以上步骤,把所得6个数字连起来,形成一串数字.求共可以形成多少种不同的数字串 思路: 网格大小只有5*5,用穷举法,不会超时. 这里利用了stl中的set容器来防止重复.最终只要输出set的size就是结果. 注意dfs回溯时的编程规范 #include<iostream> #include<algorithm> #include<s…

http://poj.org/problem?id=3050 给定一个5×5矩阵,问选6个数的不同排列总数是多少! 二维的搜索,注意要判重,数据量很小,直接用map就好. #include<cstdio> #include<cstring> #include<string> #include<iostream> #include<map> using namespace std; map<string,int>m; ][],ans;…

http://poj.org/problem?id=3050 题目大意: 在一个5*5的格子中走,每一个格子有个数值,每次能够往上下左右走一格,问走了5次后得到的6个数的序列一共同拥有多少种?(一開始站的位置算一个,能够走回去) 思路: 近期我就是在做水题... 直接DFS就可以..我用map推断序列是否反复的. #include<cstdio> #include<cstring> #include<map> #include<string> #includ…

The cows play the child's game of hopscotch in a non-traditional way. Instead of a linear set of numbered boxes into which to hop, the cows create a 5x5 rectilinear grid of digits parallel to the x and y axes.They then adroitly hop onto any digit in…

Hopscotch Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 2774 Accepted: 1940 Description The cows play the child's game of hopscotch in a non-traditional way. Instead of a linear set of numbered boxes into which to hop, the cows create a…

Hopscotch Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6761   Accepted: 4354 Description The cows play the child's game of hopscotch in a non-traditional way. Instead of a linear set of numbered boxes into which to hop, the cows creat…

用stack保存数字,set判重.dfs一遍就好.(或者编码成int,快排+unique #include<cstdio> #include<iostream> #include<string> #include<cstring> #include<queue> #include<vector> #include<stack> #include<vector> #include<map> #incl…

Hopscotch Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 2845   Accepted: 1995 Description The cows play the child's game of hopscotch in a non-traditional way. Instead of a linear set of numbered boxes into which to hop, the cows creat…

-->Hopscotch 这接写中文了 Descriptions: 奶牛们以一种独特的方式玩孩子们的跳房子游戏. 奶牛们创造了一个5x5的格子 他们熟练地跳上其中的一个格子,可以前后左右地跳(不能对角)到另一个格子上.之后继续跳(可能跳到曾经跳过的格子上). 他们总共跳5次,路径可以看作一个六位数 (准确的说是一个六位序列,如000201是可行的). 请你找到这样的六位序列的总数 Input * 输入一个5x5的地图 Output * 所有可能六位序列的总数 Sample Input 1 1 1…

Hopscotch Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 2506   Accepted: 1784 Description The cows play the child's game of hopscotch in a non-traditional way. Instead of a linear set of numbered boxes into which to hop, the cows creat…

跳格子 题目大意:牛像我们一样跳格子,一个5*5的方格,方格有数字,给牛跳5次,可以组成一个6个数字组合字符串,请问能组合多少个字符串? 题目规模很小,暴力枚举,然后用map这个玩具来检测存不存在就可以了,水题 #include <iostream> #include <functional> #include <algorithm> #include <string> #include <map> using namespace std; ][…

Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The excitement takes place on a long, straight river with a rock at the start and another rock at the end, L uni…

题意:给定一个5*5的地图,每个格子上有一个数字.从一个格子出发(上下左右4个方向),走5步将数字连起来可以构造出一个6位数.问该地图可以构造出多少个不同的6位数. 分析:可以对每个格子做深度优先遍历,构造出所有数字,但要注意不要重复计数.在这里,我使用了set来保存已构造出的数字,结果就是set中的元素个数. #include <cstdio> #include <set> #include <algorithm> using namespace std; //输入…

题意:一个5*5地图上面,从任意位置上下左右跳五次,组成一个数.问:不重复的数有多少个? 思路:dfs 从任意位置跳5次,说明每个位置都需要遍历. 组成一个数:number*10+map[dx][dy] 不重复的数字,用set(集合)来存储 只需要每次跳的时候步数加1,并且可以跳的位置,只要不超过范围就可以,即一个位置可以重复跳 解决问题的代码: #include <iostream> #include <cstdio> #include <set> using nam…

思路: 枚举+搜一下+判个重 ==AC //By SiriusRen #include <set> #include <cstdio> using namespace std; int a[8][8],xx[]={1,-1,0,0},yy[]={0,0,1,-1}; set<int>s; bool check(int x,int y){ return x>0&&x<6&&y>0&&y<6; } vo…

POJ 1852 Ants POJ 2386 Lake Counting POJ 1979 Red and Black AOJ 0118 Property Distribution AOJ 0333 Ball POJ 3009 Curling 2.0 AOJ 0558 Cheese POJ 3669 Meteor Shower AOJ 0121 Seven Puzzle POJ 2718 Smallest Difference POJ 3187 Backward Digit Sums POJ 3…

参考:http://www.cplusplus.com/reference/set/set/ 一.set 是按特定顺序存储唯一元素的容器 实现是一种非常高效的平衡检索二叉树:红黑树(Red-Black Tree). 二.set 的特性 1.set中的元素都是排好序的(与lower_bound()等结合使用能起到找前驱.后继的作用) 2.set集合中没有重复的元素(常常用于去重) 三.set 的成员函数 begin() 返回指向第一个元素的迭代器 end() 返回指向最后一个元素的迭代器 // s…

A.Lake Counting(POJ 2386) 题意: 由于最近的降雨,农夫约翰田地的各个地方都有水汇聚,用N x M(1 <= N <= 100; 1 <= M <= 100)的矩形表示.每个方格包含水('W')或干燥土地('.').农夫约翰想弄清楚他的田地里形成了多少个池塘.池塘是一组相连的正方形,里面有水,其中一个正方形被认为与八个池塘相邻给定农夫约翰的田野图,确定他有多少个池塘. 思路: 染色法,移动可用(-1,0)类似数组,也可以用for循环,注意循环时递归可能会比在…

Description Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The excitement takes place on a long, straight river with a rock at the start and another rock at th…

题目传送门 /* 二分:搜索距离,判断时距离小于d的石头拿掉 */ #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> using namespace std; typedef long long ll; ; const int INF = 0x3f3f3f3f; ll a[MAXN]; int n, m; bool check(ll d) { ; ; ; i&…

E - River Hopscotch POJ - 3258 Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The excitement takes place on a long, straight river with a rock at the start and…

River Hopscotch Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 11031   Accepted: 4737 Description Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river.…

River Hopscotch Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 5193 Accepted: 2260 Description Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The e…

题目:http://poj.org/problem?id=3258 题意: 一条河长度为 L,河的起点(Start)和终点(End)分别有2块石头,S到E的距离就是L. 河中有n块石头,每块石头到S都有唯一的距离 问现在要移除m块石头(S和E除外),每次移除的是与当前最短距离相关联的石头, 要求移除m块石头后,使得那时的最短距离尽可能大,输出那个最短距离. 和3273差不多... #include <iostream> #include <cstdio> #include <…

题目:http://poj.org/problem?id=3258 又A一道,睡觉去了.. #include <stdio.h> #include <algorithm> ]; int s, n, m; bool judge(int mid) { , cnt = ; ; i <= n+; i++) { sum += d[i] - d[i-]; if(sum < mid) cnt++; else sum = ; } if(cnt > m) ; ; } int mai…

Description Every year the cows hold an ≤ L ≤ ,,,). Along the river between the starting and ending rocks, N ( ≤ N ≤ ,) more rocks appear, each at an integral distance Di < Di < L). To play the game, each cow in turn starts at the starting rock and…

题目真是不好读,大意例如以下(知道题意就非常好解了) 大致题意: 一条河长度为 L,河的起点(Start)和终点(End)分别有2块石头,S到E的距离就是L. 河中有n块石头,每块石头到S都有唯一的距离 问如今要移除m块石头(S和E除外),每次移除的是与当前最短距离相关联的石头,要求移除m块石头后,使得那时的最短距离尽可能大,输出那个最短距离. //Memory Time //420K 391MS #include<iostream> #include<algorithm> usi…

Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The excitement takes place on a long, straight river with a rock at the start and another rock at the end, L uni…

River Hopscotch Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 21939 Accepted: 9081 Description Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The…

River Hopscotch Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 6697   Accepted: 2893 Description Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. T…