DFS(递归)模板 ——八皇后和迷宫寻找出路

Posted 2021-03-05 asdfknjhu

八皇后代码 来自 https://www.bilibili.com/video/av21776496?from=search&seid=14795429927506117804

迷宫寻路自己写的

 

迷宫寻路（1 为障碍，2 为路）

#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<stdlib.h>
#define N 1234
int n, m, count = 0;
int map[N][N];
int dx[4] = { -1,0,1,0 }, dy[4] = { 0,1,0,-1 };
typedef struct Nodes
{
    int x;
    int y;
}st; st way[N];
void out()
{
    printf("(0,0)->");
    for (int i = 0; i < count - 1; i++)
    {
        printf("(%d,%d)->", way[i].x, way[i].y);
    }
    printf("(%d,%d)
", way[count - 1].x, way[count - 1].y);
}
void DFS(int a, int b)
{
    if (a == m - 1 && b == n - 1)
    {
        out();
        return;
    }
    for (int i = 0; i < 4; i++)
    {
        if (a + dx[i] < 0 || b + dy[i] < 0 || a + dx[i] >= m || b + dy[i] >= n)
            continue;
        if (map[a + dx[i]][b + dy[i]] == 0)
        {
            map[a][b] = 1;
            way[count].x = a + dx[i];
            way[count++].y = b + dy[i];

            DFS(a + dx[i], b + dy[i]);

            count--;
            map[a][b] = 0;
        }
    }
}
// 入口是 map[0][0],出口是 map[m-1][n-1]
int main(void)
{
    scanf("%d%d", &m, &n);
    for (int i = 0; i < m; i++)
    {
        for (int j = 0; j < n; j++)
        {
            scanf("%d", &map[i][j]);
        }
    }
    DFS(0, 0);

    system("pause");
    return 0;
}
/*
测试数据
第一组
6 5
0 0 1 1 1
0 0 0 0 1
1 0 1 0 1
1 0 1 0 1
1 1 1 0 1
1 0 1 0 0
结果
(0,0)->(0,1)->(1,1)->(1,2)->(1,3)->(2,3)->(3,3)->(4,3)->(5,3)->(5,4)
(0,0)->(1,0)->(1,1)->(1,2)->(1,3)->(2,3)->(3,3)->(4,3)->(5,3)->(5,4)

第二组
6 5
0 0 1 1 1
0 0 0 0 1
1 0 1 0 1
1 0 1 0 1
1 0 1 0 1
1 0 0 0 0
结果
(0,0)->(0,1)->(1,1)->(1,2)->(1,3)->(2,3)->(3,3)->(4,3)->(5,3)->(5,4)
(0,0)->(0,1)->(1,1)->(2,1)->(3,1)->(4,1)->(5,1)->(5,2)->(5,3)->(5,4)
(0,0)->(1,0)->(1,1)->(1,2)->(1,3)->(2,3)->(3,3)->(4,3)->(5,3)->(5,4)
(0,0)->(1,0)->(1,1)->(2,1)->(3,1)->(4,1)->(5,1)->(5,2)->(5,3)->(5,4)

*/

View Code

八皇后

#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<stdlib.h>
int map[10];
int visl[10];
void out()
{
    for (int i = 1; i < 8; i++)
    {
        for (int j = i + 1; j <= 8; j++)
        {
            if (i - j == map[i] - map[j] || i - j == map[j] - map[i])  // 斜线
                return;
        }
    }
    for (int i = 1; i <= 8; i++)
    {
        printf("(%d , %d) ", i, map[i]);
    }puts("");
}
void dfs(int num)
{
    if (num >= 9)
    {
        out();
        return;
    }
    for (int i = 1; i <= 8; i++)
    {
        if (visl[i] == 0)
        {
            visl[i] = 1;
            map[num] = i;

            dfs(num + 1);

            visl[i] = 0;
        }
    }
}
int main(void)
{
    dfs(1);

    system("pause");
    return 0;
}

View Code

 

就将这两题作为模板吧。

两者都是需要记录所有路径，在根据结束条件判断是否输出

 

所以这种模板的功能是：

① 记录所有路径

② 可以根据结束条件，选择你所需要的路径

 

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梦想是注定孤独的旅行，路上少不了质疑和嘲笑，但那又怎样，哪怕遍体鳞伤，也要活的漂亮