poj 2488 A Knight's Journey ——dfs
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题目:
E - A Knight's Journey
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Submit Status
Description
Background
The knight is getting bored of seeing the same black and white squares again and again and has decided to make a journey
around the world. Whenever a knight moves, it is two squares in one direction and one square perpendicular to this. The world
of a knight is the chessboard he is living on. Our knight lives on a chessboard that has a smaller area than a regular 8 * 8 board,
but it is still rectangular. Can you help this adventurous knight to make travel plans?
Problem
Find a path such that the knight visits every square once. The knight can start and end on any square of the board.
Input
The input begins with a positive integer n in the first line. The following lines contain n test cases. Each test case consists of a
single line with two positive integers p and q, such that 1 <= p * q <= 26. This represents a p * q chessboard, where p describes
how many different square numbers 1, . . . , p exist, q describes how many different square letters exist. These are the first q
letters of the Latin alphabet: A, . . .
Output
The output for every scenario begins with a line containing "Scenario #i:", where i is the number of the scenario starting at 1.
Then print a single line containing the lexicographically first path that visits all squares of the chessboard with knight moves followed
by an empty line. The path should be given on a single line by concatenating the names of the visited squares. Each square name
consists of a capital letter followed by a number.
If no such path exist, you should output impossible on a single line.
Sample Input
3
1 1
2 3
4 3
Sample Output
Scenario #1:
A1
Scenario #2:
impossible
Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4
题目大意:
给一个最大26*26的图,只能按照国际象棋皇后的走法,求从(1,1)出发能否走完全图的每一个点.并打印可能路径
题目思路:
因为要求打印结果路径,所以需要用ans【】数组保留结果,一旦到达成立,就可以打印结果。搜索时按照字典序来搜索,dfs。成立的结果是长度为m*n
程序:
#include<iostream>#include<cstdio>#include<string>#include<cstring>#include<cmath>#include<algorithm>#include<cctype>using namespace std;int fang[8][2]= { -2,-1, -2,1, -1,-2, -1,2, 1,-2, 1,2, 2,-1, 2,1};bool v[30][30];struct la{ int x,y;}a[900];int dfs(int m,int n,int xx,int yy,int bu){ //cout<<xx<<yy<<bu<<endl; a[bu].x=xx,a[bu].y=yy; if(bu==m*n) return 1; for(int i=0; i<8; i++) { int x=xx+fang[i][0]; int y=yy+fang[i][1]; if(x>0&&y>0&&x<=m&&y<=n) if(!v[x][y]) { v[x][y]=1; if(dfs(m,n,x,y,bu+1)) return 1; v[x][y]=0; } } return 0;}int main(){ int ci,m,n,t=1; scanf("%d",&ci); while(ci--) { memset(v,0,sizeof(v)); scanf("%d%d",&n,&m); v[1][1]=1; printf("Scenario #%d:\n",t++); if(dfs(m,n,1,1,1)) { for(int i=1;i<=m*n;i++) printf("%c%d",a[i].x+'A'-1,a[i].y); printf("\n"); } else printf("impossible\n"); if(ci) printf("\n"); } return 0;}
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