搜索 || DFS || POJ 2488 A Knight's Journey

Posted 2020-10-23 舒羽倾

给一个矩形棋盘，每次走日字，问能否不重复的走完棋盘的每个点，并将路径按字典序输出

*解法：按字典序输出路径，因此方向向量的数组按字典序写顺序，dfs+回溯，注意flag退出递归的判断，并且用pre记录路径

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
char a[30][30];
int dx[] = {-2, -2, -1, -1, 1, 1, 2, 2};
int dy[] = {-1, 1, -2, 2, -2, 2, -1, 1};
int vis[30][30];
int P, Q, flag = 0;
struct node
{
    int x, y;
}pre[30][30];
void dfs(int x, int y, int step)
{
    if(step == P * Q)
    {
        flag = 1;
        return;
    }
    if(flag) return;//全走完了之后就不用再走了
    for(int i = 0; i < 8; i++)
    {
        int xx = x + dx[i], yy = y + dy[i];
        if(xx >= 0 && xx < P && yy >= 0 && yy < Q && !vis[xx][yy])
        {
            vis[xx][yy] = 1;
            step++;
            pre[x][y] = (node){xx, yy};//用pre输出路径
            dfs(xx, yy, step);
            if(flag) return;//===全走完了之后就不用再走了===
            vis[xx][yy] = 0;
            step--;
        }
    }
    return;
}
int main()
{
    int T, cc = 0;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d %d", &Q, &P);
        memset(vis, 0, sizeof(vis));
        flag = 0;
        vis[0][0] = 1;
        dfs(0,0,1);
        printf("Scenario #%d:\n", ++cc);
        if(!flag) printf("impossible\n");
        else
        {
            int ux = 0, uy = 0, vx, vy;
            for(int i = 0; i < P * Q; i++)
            {
                char c = ux + ‘A‘;
                int z = uy + 1;
                printf("%c%d", c, z);
                vx = pre[ux][uy].x, vy = pre[ux][uy].y;
                ux = vx, uy = vy;
            }
            printf("\n");
        }
        printf("\n");
    }
    return 0;
}