poj 2488 DFS
2012-07-25 21:35
288 查看
A Knight's Journey
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other)Total Submission(s) : 10 Accepted Submission(s) : 4
[align=left]Problem Description[/align]
View Code
#include <iostream> #include <stdio.h> #include <string.h> using namespace std; int vis[26][26],path[26][2]; int flag,a,b; int dir[8][2]={-2,-1,-2,1,-1,-2,-1,2, //这样子排数字是为了保证字典序 1,-2, 1,2, 2,-1, 2,1}; void dfs(int s,int t, int k) { int i,j; if(k==a*b) { for(i=0;i<k;i++) printf("%c%d",path[i][0]+'A',path[i][1]+1); printf("\n"); flag=1; } else { for(j=0;j<8;j++) { int n=s+dir[j][0]; int m=t+dir[j][1]; if(n>=0 && n<b && m>=0 && m<a && vis [m]==0 && !flag) { vis [m]=1; path[k][0]=n; path[k][1]=m; dfs(n,m,k+1); vis [m]=0; } } } } int main() { int n,t; scanf("%d",&n); for(t=1;t<=n;t++) { flag=0; int i,j; scanf("%d%d",&a,&b); for(i=0;i<a;i++) { for(j=0;j<b;j++) { vis[i][j]=0; } } vis[0][0]=1; path[0][0]=0;path[0][1]=0; printf("Scenario #%d:\n",t); dfs(0,0,1); if(flag==0) printf("impossible\n"); printf("\n"); } //system("pause"); return 0; }
相关文章推荐
- poj 2488 A Knight's Journey 【骑士周游 dfs + 记忆路径】
- poj 2488 A Knight's Journey (dfs)
- poj 2488 A Knight's Journey 【dfs】【字典序】【刷题计划】
- poj2488 A Knight's Journey 简单DFS 注意搜索步骤
- POJ 2488 A Knight's Journey (DFS)
- POJ 2488 A Knight's Journey (dfs+改变搜索顺序)
- POJ 2488 - A Knight's Journey (DFS)
- poj 2488 -- A Knight's Journey (DFS)
- POJ 2488 A Knight's Journey(DFS)
- POJ 2488 A Knight's Journey DFS深搜
- POJ 2488 DFS 模拟 马的跳动
- poj 2488 A Knight's Journey(暴力dfs 你懂的)
- POJ 2488 A Knight's Journey【dfs】
- POJ 2488 A Knight's Journey【DFS】
- POJ 2488 A Knight's Journey (DFS + 记录路径)
- POJ 2488 简单 DFS
- POJ 2488-A Knight's Journey(dfs)
- poj 2488 dfs+回溯
- POJ 2488 A Knight's Journey (dfs)
- POJ 2488 DFS