poj3009Curling 2.0(简单dfs)

->题目请戳这里<-

题目大意：在一个h*w的场地内玩冰壶游戏，0代表空地，1代表障碍，2代表冰壶起始位置，3代表目标位置。现在要把冰壶挪到目标位置，冰壶只能往上下左右4个方向走，冰壶隔壁如果是障碍物，就不能往该方向走。冰壶走了就不会停下来，除非碰到障碍物，碰到障碍物后冰壶停止，障碍物消失（先停止再消失），冰壶当然不能出界。求冰壶能否在10步内（包括10步）到达目的地，如果能，输出最小步数，不能输出-1。

题目分析：题目要求最小步数，考虑到场地最大20*20，并且步数限制不超过10步，考虑用dfs回溯遍历所有解，取最小的。注意输入的时候是先输入列，再输入行，很条件反射的先处理行再处理列，提交上去果断WA，还sb的调了半天，测了n组数据才发现>o<。。。。中招的童鞋请自行了断。。。

详情请见代码：

#include <iostream>#include<cstdio>#include<cstring>using namespace std;const int N = 30;int map[N][N];int h,w;int si,sj,ei,ej;struct node{    int x,y;}path[20];int ans;int dir[][2] = {{1,0},{-1,0},{0,1},{0,-1}};int isok(int x,int y){    return (x > 0 && y > 0 && x <= h && y <= w);}void dfs(int curi,int curj,int step){    path[step].x = curi;    path[step].y = curj;    if(step > 10)        return;    if(curi == ei && curj == ej)    {        if(step < ans)            ans = step;        //printf("path:%d\n",step);        //for(int i = 0;i <= step;i ++)        //    printf("%d %d\n",path[i].x,path[i].y);        return;    }    int ti,tj,i;    for(i = 0;i < 4;i ++)    {        ti = curi;        tj = curj;        if(map[ti + dir[i][0]][tj + dir[i][1]] != 1)//要走的方向上相邻的不是障碍，说明能走        {            while(isok(ti,tj) && map[ti][tj] == 0)            {                ti += dir[i][0];                tj += dir[i][1];            }            if(map[ti][tj] == 1)//            {                map[ti][tj] = 0;                dfs(ti - dir[i][0],tj - dir[i][1],step + 1);                map[ti][tj] = 1;            }            if(map[ti][tj] == 3)                dfs(ti,tj,step + 1);        }    }}int main(){    int i,j;    while(scanf("%d%d",&w,&h),(h + w))    {        memset(map,0,sizeof(map));        for(i = 1;i <= h;i ++)        {            for(j = 1;j <= w;j ++)            {                scanf("%d",&map[i][j]);                if(map[i][j] == 2)                {                    si = i;                    sj = j;                }                else                {                    if(map[i][j] == 3)                    {                        ei = i;                        ej = j;                    }                }            }        }        ans = 10000;        map[si][sj] = 0;        //printf("si:%d  sj:%d\n",si,sj);        dfs(si,sj,0);        //printf("ans:");        if(ans > 10)            printf("-1\n");        else            printf("%d\n",ans);    }    return 0;}/*2 13 26 61 0 0 2 1 01 1 0 0 0 00 0 0 0 0 30 0 0 0 0 01 0 0 0 0 10 1 1 1 1 16 11 1 2 1 1 36 11 0 2 1 1 312 12 0 1 1 1 1 1 1 1 1 1 313 12 0 1 1 1 1 1 1 1 1 1 1 30 0*//*18 141 0 1 0 1 1 1 0 0 0 0 1 1 1 0 1 0 30 1 0 1 0 1 1 0 0 0 0 1 1 0 1 1 1 00 0 0 0 0 1 0 0 1 1 0 0 0 0 1 0 1 01 1 0 1 1 1 1 1 1 0 1 1 1 1 0 1 1 01 1 1 0 0 1 1 0 0 2 1 1 1 1 1 1 0 10 1 1 0 0 0 0 0 1 1 1 1 0 0 0 1 1 01 0 0 1 1 1 0 1 0 1 0 0 0 0 1 0 1 00 1 1 1 0 1 1 1 0 0 1 0 1 0 0 0 0 00 1 1 1 0 1 0 1 1 1 0 1 1 1 1 1 0 01 0 0 1 1 1 0 0 1 0 1 0 1 1 1 0 0 00 1 0 0 1 0 1 0 1 0 1 1 1 1 0 1 0 11 1 1 1 0 1 1 1 0 1 0 0 0 0 1 1 1 01 0 0 0 1 1 1 0 1 1 1 1 1 0 1 0 0 00 0 1 1 0 1 0 0 0 0 1 1 1 1 0 0 1 014 120 1 0 0 1 1 0 1 0 1 0 1 2 10 0 0 0 0 0 0 0 0 1 0 0 1 11 0 0 1 1 1 1 0 0 0 0 0 1 01 0 1 1 1 0 1 0 3 1 1 0 0 00 1 0 0 1 0 1 0 1 0 0 0 1 00 0 0 1 0 1 1 1 0 1 1 0 0 01 0 1 0 0 1 0 0 0 1 0 0 0 01 0 0 1 1 1 1 1 1 0 0 0 0 00 1 0 1 0 1 0 1 1 0 0 0 0 01 1 1 1 1 0 0 1 0 0 0 1 1 10 0 1 1 1 0 1 1 1 1 1 0 0 00 0 0 1 0 0 1 1 1 0 1 0 1 09 131 0 1 1 0 0 1 0 00 0 1 1 1 1 1 0 11 1 0 1 0 0 0 0 11 1 1 0 0 0 0 1 01 1 0 1 1 1 1 1 10 1 0 1 0 1 1 0 00 1 0 3 1 1 0 0 10 1 0 1 1 0 0 0 12 1 1 0 0 1 0 0 10 1 0 1 0 0 1 1 01 0 1 1 1 0 0 1 10 0 1 0 0 1 0 0 01 1 1 0 0 1 1 1 02 73 01 10 01 01 00 12 04 131 0 1 01 0 0 11 0 1 13 1 0 01 0 1 10 1 1 00 0 0 02 1 0 11 1 0 01 1 1 11 0 0 00 0 1 01 0 0 114 190 1 0 1 1 0 0 1 1 1 1 1 0 01 0 0 0 3 0 0 1 1 1 2 0 1 10 1 1 1 0 1 0 0 0 0 1 0 0 00 1 1 0 0 0 0 1 0 0 1 0 0 00 0 1 1 0 1 1 0 1 0 0 1 0 11 0 0 1 1 1 1 1 0 1 1 0 1 01 0 0 0 1 0 0 0 1 0 1 0 0 01 0 1 0 0 1 0 0 1 0 1 0 1 00 1 0 0 0 1 0 0 0 0 1 1 0 00 1 0 0 1 1 0 1 1 1 0 1 0 11 1 0 1 1 0 0 1 0 0 0 0 1 11 0 1 1 0 1 1 1 1 1 1 1 1 00 1 1 0 0 1 0 1 1 1 1 0 0 11 1 0 1 0 1 1 0 1 1 0 0 0 01 0 1 1 0 0 0 0 0 0 0 1 0 10 1 0 0 1 0 1 0 0 1 1 0 1 01 0 1 0 0 0 1 1 0 0 1 1 1 10 1 1 0 1 1 0 0 0 1 0 1 0 11 1 0 0 1 0 1 1 0 0 0 1 0 05 41 0 0 1 03 1 1 1 21 0 0 1 11 1 0 0 018 60 1 1 0 0 1 0 0 0 0 0 1 0 1 1 1 0 01 0 1 0 1 1 0 0 1 1 1 0 0 0 1 1 1 10 1 0 0 0 1 0 0 0 1 0 1 0 1 0 1 1 11 1 0 0 0 1 0 0 0 1 1 0 1 0 1 2 0 01 0 3 1 1 1 0 0 0 0 1 1 0 0 0 0 1 11 1 0 1 1 0 1 0 1 1 0 1 0 0 1 1 1 019 171 1 1 1 0 1 0 0 1 0 0 0 0 0 1 1 0 0 00 0 1 1 0 0 0 0 0 1 0 0 0 1 0 0 0 0 10 1 0 1 1 1 0 0 1 0 1 0 1 1 1 1 1 1 01 0 1 1 0 0 1 0 0 1 0 1 0 1 1 0 0 0 01 0 1 0 0 1 1 1 1 0 0 0 1 0 1 0 0 0 01 0 0 0 1 1 1 0 1 1 1 1 1 0 1 1 1 0 01 0 0 0 0 1 1 3 1 1 0 0 0 0 1 0 0 0 00 0 1 1 0 1 0 0 0 0 1 1 0 0 1 0 1 1 00 1 1 0 0 0 1 0 1 1 1 1 0 1 0 0 0 2 11 1 1 0 1 1 0 1 0 0 1 0 0 0 0 0 1 1 00 0 0 1 0 0 0 1 1 0 1 0 1 1 1 0 0 1 00 1 0 0 0 0 0 0 1 1 1 0 0 1 1 0 1 0 11 1 0 0 0 1 0 0 0 1 1 0 0 0 1 0 0 1 10 0 1 1 0 0 0 0 1 1 0 0 0 1 0 0 1 1 01 0 0 0 1 1 0 1 1 1 0 0 1 1 1 1 0 1 00 1 0 1 1 1 0 0 1 0 1 0 1 1 0 1 0 1 11 0 1 0 0 0 1 1 1 1 1 1 0 1 1 1 0 0 018 21 1 0 1 0 1 1 0 1 0 0 3 0 0 1 0 1 00 0 0 1 0 1 0 1 1 2 1 1 0 1 0 0 1 07 101 0 0 0 1 1 01 0 1 1 1 0 10 1 1 0 1 1 10 1 1 1 0 1 11 1 0 0 0 1 01 1 0 1 2 1 00 3 0 0 1 0 00 1 1 1 0 0 01 0 1 0 1 0 01 1 1 1 1 1 014 60 0 1 0 0 0 0 1 1 0 1 1 1 00 0 1 0 1 0 1 1 1 1 1 1 1 11 1 0 2 1 1 0 0 1 0 0 0 0 11 0 1 1 1 1 1 1 1 1 1 1 1 01 0 1 1 1 0 1 1 0 1 1 0 0 01 0 1 0 0 1 0 0 1 0 0 0 0 34 80 1 0 00 0 0 01 0 0 10 0 0 00 0 0 03 0 1 01 0 1 01 0 0 2*///136K297MS