Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 28697 | Accepted: 9822 |
Description
Input
Output
Sample Input
3 1 1 2 3 4 3
Sample Output
Scenario #1: A1 Scenario #2: impossible Scenario #3: A1B3C1A2B4C2A3B1C3A4B2C4
#include <cstdio> #include <cstdlib> #include <stack> #include <cstring> using namespace std; const int MAX = 9; const int dirx[8]={-1,1,-2,2,-2,2,-1,1},diry[8]={-2,-2,-1,-1,1,1,2,2}; typedef struct Point{ int x,y; }point; int p,q,n; bool visit[MAX][MAX]; point pre[MAX][MAX]; bool mark; stack<int> stx,sty; void printPath(int x,int y){ stx.push(x); sty.push(y); int tx,ty; tx = pre[x][y].x; ty = pre[x][y].y; while(tx!=-1){ stx.push(tx); sty.push(ty); x = pre[tx][ty].x; y = pre[tx][ty].y; tx = x; ty = y; } while(!stx.empty()){ printf("%c%d",sty.top()-1+‘A‘,stx.top()); stx.pop(); sty.pop(); } printf("\n\n"); } void dfs(int x,int y,int len){ if(mark)return; if(len==p*q){ printPath(x,y); mark = true; return; } int i,tx,ty; for(i=0;i<8;++i){ tx = x+dirx[i]; ty = y+diry[i]; if(tx<1 || tx>p || ty<1 || ty>q)continue; if(visit[tx][ty])continue; pre[tx][ty].x = x; pre[tx][ty].y = y; visit[tx][ty] = true; dfs(tx,ty,len+1); visit[tx][ty] = false; } } int main() { //freopen("in.txt","r",stdin); //(Author : CSDN iaccepted) int i; scanf("%d",&n); for(i=1;i<=n;++i){ printf("Scenario #%d:\n",i); scanf("%d %d",&p,&q); memset(visit,0,sizeof(visit)); mark = false; pre[1][1].x = -1; pre[1][1].y = -1; visit[1][1] = true; dfs(1,1,1); visit[1][1] = false; if(!mark){ printf("impossible\n\n"); } } return 0; }
思路就是DFS 搜下去,当走过的格子数达到格子总数时就打印路径。所以要用一个数组记录每个定点的前驱节点。
pku 2488 A Knight's Journey (搜索 DFS),布布扣,bubuko.com
pku 2488 A Knight's Journey (搜索 DFS)
原文地址:http://blog.csdn.net/iaccepted/article/details/26341633