HDU1372:Knight Moves(经典BFS题),hdu1372bfs


                                          HDU1372:Knight Moves(BFS)

3000MS     Memory Limit:0KB     64bit IO Format:%lld &


A friend of you is doing research on the Traveling Knight Problem
 where you are to find the shortest closed tour of knight moves
that visits each square of a given set of n squares on a chessboard
exactly once. He thinks that the most difficult part of the problem is
determining the smallest number of knight moves between two given
squares and that, once you have accomplished this, finding the tour
would be easy.

Of course you know that it is vice versa. So you offer him to write a
program that solves the "difficult" part.


Your job is to write a program that takes two squares a and b as
input and then determines the number of knight moves on a shortest route
from a to b.


A Math Problem

Input Specification

The input file will contain one or more test cases. Each test case
consists of one line containing two squares separated by one space. A
square is a string consisting of a letter (a-h) representing the
column and a digit (1-8) representing the row on the chessboard.


Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K
Total Submission(s): 1867 Accepted Submission(s): 519

Output Specification

For each test case, print one line saying
"To get from xx to yy takes n knight moves.".


Problem Description
You are given a positive integer n, please count how many positive
integers k satisfy kk≤n.

Sample Input


e2 e4a1 b2b2 c3a1 h8a1 h7h8 a1b1 c3f6 f6


There are no more than 50 test cases.

Sample Output


To get from e2 to e4 takes 2 knight moves.To get from a1 to b2 takes 4 knight moves.To get from b2 to c3 takes 2 knight moves.To get from a1 to h8 takes 6 knight moves.To get from a1 to h7 takes 5 knight moves.To get from h8 to a1 takes 6 knight moves.To get from b1 to c3 takes 1 knight moves.To get from f6 to f6 takes 0 knight moves.




#include <iostream>#include <queue>#include<string.h>using namespace std;int s1,s2,e1,e2;int tu[8][8];int xx[8] = {1, 2, 1, 2, -1, -2, -1, -2};//x坐标变化int yy[8] = {2, 1, -2, -1, 2, 1, -2, -1};//y坐标变化int bfs(){ memset(tu,0,sizeof(tu));//数组值全为0 queue<int> q; int x1,y1,x2,y2; tu[s1][s2]=0;// q.push(s1); q.push(s2); while(!q.empty()) { x1=q.front(); q.pop(); y1=q.front(); q.pop(); if(x1==e1&&y1==e2)//在原地不动 return tu[x1][y1]; for(int i=0; i<8; i++) { x2=x1+xx[i]; y2=y1+yy[i]; if(x2<0||x2>7||y2<0||y2>7||tu[x2][y2]>0) continue; tu[x2][y2]=tu[x1][y1]+1; q.push(x2); q.push(y2); } } return 0;}int main(){ char a[3],b[3]; int total; while(cin>>a>>b) { s1=a[0]-'a';//输入的字母,是列,从第一列a开始,减去'a',转换 s2=a[1]-'1';//输入的数字,是行,从的第一行开始,减去'1',转换 e1=b[0]-'a'; e2=b[1]-'1'; total=bfs(); cout<<"To get from "<<a<<" to "<<b<<" takes "<<total<<" knight moves."<<endl; } return 0;}


HDU1372:Knight Moves(BFS) Time Limit: 3000MS Memory Limit: 0KB 64bit IO
Format: %lld %llu Description A friend of you is doing resea...

Each case only contains a positivse integer n in a line.


For each test case, output an integer indicates the number of positive
integers k satisfy kk≤n in a line.

Sample Input

Sample Output

#include <iostream>#include <cstring>#define LL long long;LL kpow(int x, int n){ LL res = 1; while(n >= 0){ if(n & 1) res = res * x; x = x * x; n >>= 1; } return res;}int main(){ LL n; while(~scanf("%lld",&n)){ for(int k = 15; k >= 1; k--){ if(kpow(k,k) <= n){ printf("%d\n",k); break; } } } return 0;}


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