//首先把每个在能走点走出去的最小步数求出并保存//在进行dfs时，求h()时，即是求当前存在的节点走出去的最小步数中最大者//其他地方和IDA*题一样处理即可，此题就是多了个预处理，首先在记忆化搜索一遍，然后在IDA*#include<stdio.h>#include<string.h>#include<algorithm>#include<queue>#define inf 12345678using namespace std;int n,m,t;char map[9][9];char dir[5][10]={"east","north","south","west"};int dx[4]={0,-1,1,0},dis[9][9];int dy[4]={1,0,0,-1};int path[1000];struct TPoint{ int x,y; TPoint(){} TPoint(int _x,int _y):x(_x),y(_y){}}st,en;queue<TPoint>tpl;void addedge(int u,int v,int val){ dis[u][v]=val; tpl.push(TPoint(u,v));}bool ismap(int x,int y){ return x>=0&&x<n&&y>=0&&y<n;}bool onedge(int x,int y){ return x==0||y==0||x==n-1||y==n-1;}void bfs(){ int i; while(!tpl.empty()) { st=tpl.front(),tpl.pop(); for(i=0;i<4;i++) { en.x=st.x+dx[i],en.y=st.y+dy[i]; if(!ismap(en.x,en.y)) continue; if(!map[en.x][en.y]) continue; if(dis[en.x][en.y]>dis[st.x][st.y]+1) { dis[en.x][en.y]=dis[st.x][st.y]+1; tpl.push(en); } } }}void init(){ int i,j; while(!tpl.empty()) tpl.pop(); for(i=0;i<n;i++) for(j=0;j<n;j++) { dis[i][j]=inf; map[i][j]=map[i][j]=='1'?0:1; if(!map[i][j]) continue; if(onedge(i,j)) addedge(i,j,0); } bfs();}int h(char mid[9][9]){ int i,j,maxf=0; for(i=0;i<n;i++) { for(j=0;j<n;j++) { if(mid[i][j]) maxf=max(maxf,dis[i][j]); } } return maxf;}bool dfs(char mid[9][9],int dep){ if(dep+h(mid)>t) return 0; if(dep==t) return 1; int i,j,k,l,r; char cvs[9][9]; for(i=0;i<4;i++) { memset(cvs,0,sizeof(cvs)); for(j=0;j<n;j++) for(k=0;k<n;k++) { if(onedge(j,k)||!mid[j][k]) continue; l=j+dx[i],r=k+dy[i]; if(!map[l][r]) cvs[j][k]=1; else cvs[l][r]=1; } path[dep]=i; if(dfs(cvs,dep+1)) return 1; } return 0;}int main(){ int i,end=0; while(scanf("%d",&n)!=EOF) { if(end++) puts(""); for(i=0;i<n;i++) scanf("%s",map[i]); init(); for(t=0;;t++) { if(dfs(map,0)) break; } for(i=0;i<t;i++) printf("%s\n",dir[path[i]]); } return 0;}
