POJ-2488-A Knight's Journey

DFS题

代码：

#include<cstdio>#include<cstring>#include<iostream>using namespace std;int p,q,map[9][9],use[9][9],ansx[9*9],ansy[9*9],ans,movex[8]={-1,1,-2,2,-2,2,-1,1},movey[8]={-2,-2,-1,-1,1,1,2,2};bool isborder(int x,int y){    if(x>0&&x<=p&&y>0&&y<=q)return false;    return true;}bool DFS(int x,int y,int res){    if(!res)    {return true;    }    for(int i=0;i<8;i++)    {int itx=x+movex[i];int ity=y+movey[i];if(isborder(itx,ity)||use[itx][ity])    continue;use[itx][ity]=1;if(DFS(itx,ity,res-1)){    ansx[ans]=itx;    ansy[ans++]=ity;    return true;}use[itx][ity]=0;    }    return false;}int main(){    int cas,cass=1;    scanf("%d",&cas);    while(cas--)    {ans=0;scanf("%d%d",&p,&q);int flag=0;for(int i=1;i<=p;i++){    for(int j=1;j<=q;j++)    {memset(use,0,sizeof(use));use[i][j]=1;if(DFS(i,j,p*q-1)){    ansx[ans]=i;    ansy[ans++]=j;    flag=1;    break;}    }    if(flag)break;}printf("Scenario #%d:\n",cass++);if(!flag)    printf("impossible\n");else{    for(int i=ans-1;i>=0;i--)printf("%c%d",ansy[i]+'A'-1,ansx[i]);    printf("\n");}printf("\n");    }    return 0;}