在横坐标上 从n点走到k点 至少要几步 可以到 n+1 n-1 n*2这3个点 Sample Input5 17 Sample Output4 #include <iostream> #include <cstring> #include <cstdio> #include <queue> using namespace std; int n , k ; int ans ; ] ; struct node { int x ; int step ; }; bo…

HDU 2717 题目大意:在x坐标上,农夫在n,牛在k.农夫每次可以移动到n-1, n+1, n*2的点.求最少到达k的步数. 思路:从起点开始,分别按x-1,x+1,2*x三个方向进行BFS,最先找到的一定是最小的步数. /* HDU 2717 Catch That Cow --- BFS */ #include <cstdio> #include <cstring> #include <queue> using namespace std; ]; int n, k…

Catch That Cow Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6383    Accepted Submission(s): 2034 Problem Description Farmer John has been informed of the location of a fugitive cow and wants…

题目链接 Problem Description Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number li…

Catch That Cow Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 8999    Accepted Submission(s): 2837 Problem Description Farmer John has been informed of the location of a fugitive cow and wants…

胜利大逃亡 Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 23779    Accepted Submission(s): 9125 Problem Description Ignatius被魔王抓走了,有一天魔王出差去了,这可是Ignatius逃亡的好机会. 魔王住在一个城堡里,城堡是一个A*B*C的立方体,可以被表示成A个B*C的矩…

n个点 m条路 询问T次 从a点走到b点刚好k步的方案数是多少 给定一个有向图,问从A点恰好走k步(允许重复经过边)到达B点的方案数mod p的值把 给定的图转为邻接矩阵,即A(i,j)=1当且仅当存在一条边i->j.令C=A*A,那么C(i,j)=ΣA(i,k)*A(k,j),实际上就 等于从点i到点j恰好经过2条边的路径数(枚举k为中转点).类似地,C*A的第i行第j列就表示从i到j经过3条边的路径数 Sample Input4 4 // n m0 10 21 32 32 //T0 3 2…

传送门:http://acm.hdu.edu.cn/showproblem.php?pid=2717 Catch That Cow Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 20259    Accepted Submission(s): 5926 Problem Description Farmer John has been i…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2717 Catch That Cow Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 12615    Accepted Submission(s): 3902 Problem Description Farmer John has been…

看题:http://acm.hdu.edu.cn/showproblem.php?pid=2717 思路:相当于每次有三个方向,加1,减1,乘2,要注意边界条件,减1不能小于0,乘2不能超过最大值. 然后还要注意N>=K的时候,只能减1才能到达. #include <iostream> #include <string> #include <cstdio> #include <cmath> #include <vector> #includ…