HDU 1010

来源：互联网发布：开淘宝网流程编辑：程序博客网时间：2024/06/15 12:36
这个题教会我DFS不要写成for循环的形式
#include <cstdio>#include <cstdlib>#include <cstring>#include <queue>#include <iostream>using namespace std;const int MAX = 10;int m, n, t, sx, sy, flag, vis[MAX][MAX];int dx[] = {1, -1, 0, 0};int dy[] = {0, 0, 1, -1};char map[MAX][MAX];void DFS(int x, int y, int step){    if(step > t || flag)    {        return ;    }    if (map[x][y] == 'D' && step == t)    {        flag = 1;        return;    }    if (map[x][y] == '.' || map[x][y] == 'S')    {        vis[x][y] = 1;        //AC        // if(x + 1 >= 1 && x + 1 <= m && y >= 1 && y <= n && vis[x + 1][y] ==0 )          //     DFS(x + 1, y, step + 1);        // if(x - 1 >= 1 && x - 1 <= m && y >= 1 && y <= n && vis[x - 1][y] ==0 )          //     DFS(x - 1, y, step + 1);        // if(x >= 1 && x <= m && y + 1 >= 1 && y + 1 <= n && vis[x][y + 1] ==0 )          //     DFS(x, y + 1, step + 1);        // if(x >= 1 && x <= m && y - 1 >= 1 && y - 1 <= n && vis[x][y - 1] ==0 )          //     DFS(x, y - 1, step + 1);        //TL        for (int i = 0; i < 4; i++)        {            int nx = x + dx[i];            int ny = y + dy[i];            if(nx >= 1 && nx <= m && ny >= 1 && ny <= n && vis[nx][ny] == 0)            {                DFS(nx, ny, step + 1);            }        }        vis[x][y] = 0;    }}int main(){    while (scanf("%d%d%d", &m, &n, &t), m + n + t)    {        getchar();        flag = 0;        memset(vis, 0, sizeof(vis));        for (int i = 1; i <= m; i++)        {            for (int j = 1; j <= n; j++)            {                map[i][j] = getchar();                if (map[i][j] == 'S')                 {                    sx = i;                    sy = j;                }            }            getchar();        }        DFS(sx, sy, 0);        printf(flag?"YES\n":"NO\n");    }    return 0;}
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