POJ 3009 Curling 2.0 (dfs)

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IN this question , their is a little bit change.first we need to find a far enough location ,than we should go one step further to change the unreachable block reachable , and than take step back to continue.

#include <iostream>#include <cstring>#include <cstdio>#include <vector>#include <algorithm>#include <set>#include <map>#define rep(i, n) for(i;i<n;i++)#define inf  0x7f7f7f7f7fusing namespace std;int b[101][101];int vis[25][25];   //indicate ith number is markedint n,m;int ans =11;int dx[4] = {0,1,0,-1};int dy[4] = {1,0,-1,0};bool check(int x,int y){    //cout<<"checking:"<<x<<", "<<y<<endl;    if(x<0||y<0||x>=n||y>=m) return false;    else return true;}void dfs(int x,int y,int step){    //cout<<x<<" "<<y<<" "<<step<<endl;    if(step>10) return ;    int leap = 0;    int xx,yy;    for(int i=0;i<4;i++)    {        leap = 0;        if(b[x+dx[i]][y+dy[i]] == 1) continue;        xx = x;        yy = y;        while(1)        {            xx += dx[i];            yy += dy[i];            if(!check(xx,yy))            {                leap = 1;                break;            }            if(b[xx][yy] == 1)            {                break;            }            if(b[xx][yy] == 0)            {                continue;            }            if(b[xx][yy] == 3)            {                ans = min(ans,step);                return ;            }        }        if(leap) continue;        b[xx][yy] = 0;        dfs(xx-dx[i],yy-dy[i],step+1);        b[xx][yy] = 1;    }}int main(){    while(~scanf("%d%d",&m,&n))    {        if(!n&&!m) break;        int sx,sy;        ans = 11;        memset(vis,0,sizeof(vis));        for(int i=0;i<n;i++)        {            for(int j=0;j<m;j++)            {                scanf("%d",&b[i][j]);                if(b[i][j] == 2)                {                    sx = i;                    sy = j;                    b[i][j] = 0;                }            }        }        dfs(sx,sy,0);        if(ans+1<11)  printf("%d\n",ans+1);        else printf("-1\n");    }    return 0;}