动态规划——走迷宫

SDUT OJ
J题

# include<stdio.h>
# include<string.h>

int a[30][30];
int b[100];     //用于储存路径输出
int dir[4][2]={{0,-1},{-1,0},{0,1},{1,0}};  //4个方向

int c,d,flag;

void work(int g,int h,int ii);

int main( ) //1269--走迷宫
{
    int n,m,g,h,i,j,ii;
    while(scanf("%d %d",&n,&m) == 2)
    {
        ii = flag = 0;
        memset(b,0,sizeof(b));
        memset(a,0,sizeof(a));
        for(i=1;i<=n;i++)
            for(j=1;j<=m;j++)
                scanf("%d",&a[i][j]);
        scanf("%d %d %d %d",&g,&h,&c,&d);
        work(g,h,0);
        if(!flag) printf("-1\n");
    }
    return 0;
}

void work(int g,int h,int ii)
{
    int i;
    if(g == c && h == d )    //来到终点则输出
    {
        flag = 1;
        for(i=0;i<ii;i+=2)
        {
            printf("(%d,%d)",b[i],b[i+1]);
            if(i<ii)
                printf("->");
        }
        printf("(%d,%d)\n",c,d);
        return;
    }
    if(!a[g][h])return ;    //到达边界或已走或墙壁返回
    for(i=0;i<4;i++)    //依次走4个方向
    {
        a[g][h] = 0;     //如果该方向可走,则标记已走,为0
        b[ii] = g;         //记下坐标
        b[ii+1] = h;
        work(g+dir[i][0],h+dir[i][1],ii+2);   //走下一步
        a[g][h] = 1;      //恢复
    }
}

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转载自blog.csdn.net/wjt_1025/article/details/89526625