过河卒 bfs搜索

过河卒

一共有三份代码，

分别是只会写迷宫时的、会写dfs时的、动态规划的，洛谷上对应的分数是40,60,100

 

 

迷宫写法

这份代码只是自己用来练习bfs的，运行的话只有40，数据过大会超时，还不是一般的超时，当m=n=20的时候，运行了一分钟还没结果。

 

用二维数组表示棋盘，与题目的棋盘相比，加上了外围的边界，所以坐标会有变化，比如：

 

原来坐标：

             

 

改后坐标：

             

 

 

思路与迷宫完全相同

 

代码如下：

#include <stdio.h>
#include <malloc.h>
#include <math.h>

#define MAXSIZE 30

typedef enum abc { RIGHT, DOWN } DIR;            //只有两个方向可走
typedef struct
{
    int x, y;
    DIR dir;
}Pos;

typedef struct
{
    Pos data[500];
    int top;
}Stack;

int matrix[MAXSIZE][MAXSIZE];

int main()
{
    int n, m, xhouse, yhouse, path_num = 0;
    Pos top_pos;
    scanf("%d %d %d %d", &n, &m, &xhouse, &yhouse);

    //这里参考的是迷宫的做法，将外围全部为1，题目中没有外围，所以中间可走的部分坐标会有些变化
    for (int i = 0; i <= n + 2; i++)
    {
        matrix[i][0] = 1;
        matrix[i][m + 2] = 1;
    }

    for (int j = 0; j <= m + 2; j++)
    {
        matrix[0][j] = 1;
        matrix[n + 2][j] = 1;
    }
    //把马和可能走到的位置标记为不能走
    for (int i = 1; i <= n + 1; i++)
        for (int j = 1; j <= m + 1; j++)
            if (  
                abs(i - (xhouse + 1)) == 1 && abs(j - (yhouse + 1)) == 2 ||
                abs(i - (xhouse + 1)) == 2 && abs(j - (yhouse + 1)) == 1 ||
                i == xhouse + 1 && j == yhouse + 1
               )

                matrix[i][j] = 1;

    Stack path;
    path.top = -1;

    path.data[++path.top].x = 1;
    path.data[path.top].y = 1;
    path.data[path.top].dir = RIGHT;

    while (path.top > -1)
    {
        top_pos.x = path.data[path.top].x;
        top_pos.y = path.data[path.top].y;
        top_pos.dir = path.data[path.top].dir;
        if (path.data[path.top].x == n + 1 && path.data[path.top].y == m + 1)
        {
            path_num++;
            if (--path.top > -1)
                path.data[path.top].dir++;
        }
        else
            switch (top_pos.dir)
        {
        case RIGHT:
            if (matrix[top_pos.x + 1][top_pos.y] == 0)
            {
                path.data[++path.top].x = top_pos.x + 1;
                path.data[path.top].y = top_pos.y;
                path.data[path.top].dir = RIGHT;
            }
            else
                path.data[path.top].dir++;
            break;
        case DOWN:
            if (matrix[top_pos.x][top_pos.y + 1] == 0)
            {
                path.data[++path.top].x = top_pos.x;
                path.data[path.top].y = top_pos.y + 1;
                path.data[path.top].dir = RIGHT;
            }
            else
                path.data[path.top].dir++;
            break;
        default:
            path.data[--path.top].dir++;

        }
    }
    printf("%d", path_num);
    return 0;

}

 

dfs

更新一下，还没有学会动态规划，倒是dfs用熟练了，开O2有60分了，看来dfs是一定超时了o(╥﹏╥)o

#include <stdio.h>
#include <math.h>
#define MAXSIZE 25

int matrix[MAXSIZE][MAXSIZE];
int xhorse, yhorse, xterminal, yterminal;
long long sum;

void dfs(int xstart,int ystart)
{
    if (xstart == xterminal && ystart == yterminal)
    {
        sum++;
        return;
    }
    if (matrix[xstart + 1][ystart] == 1)
        dfs(xstart + 1, ystart);
    if (matrix[xstart][ystart + 1] == 1)
        dfs(xstart, ystart + 1);
}

int main()
{
    
    scanf("%d", &xterminal);
    scanf("%d", &yterminal);
    scanf("%d", &xhorse);
    scanf("%d", &yhorse);
    
    for (int i = 0; i <= xterminal; i++)//0表示非棋盘位置，1表示可以走的位置，2表示马的位置，不能走
        for (int j = 0; j <= yterminal; j++)
            if (abs(i - xhorse) == 1 && abs(j - yhorse) == 2 || abs(i - xhorse) == 2 && abs(j - yhorse) == 1|| i == xhorse && j == yhorse)
            {
                matrix[i][j] = 2;
            }
            else
                matrix[i][j] = 1;

    dfs(0, 0);
    printf("%lld", sum);
    return 0;
}

 

动态规划

最终还是放弃了dfs，改用动态规划了，100分过了，但还是对dfs不死心啊

#include <stdio.h>
#include <math.h>

#define MAXSIZE 25

long long matrix[MAXSIZE][MAXSIZE];

int xhorse, yhorse, xterminal, yterminal;

int main()
{
    scanf("%d", &xterminal);
    scanf("%d", &yterminal);
    scanf("%d", &xhorse);
    scanf("%d", &yhorse);
    
    for (int i = xterminal; i >= 0; i--)
        for (int j = yterminal; j >= 0; j--)
        {
            if (i == xterminal && j == yterminal)
                matrix[i][j] = 1;
            else if (abs(i - xhorse) == 1 && abs(j - yhorse) == 2 || abs(i - xhorse) == 2 && abs(j - yhorse) == 1 || i == xhorse && j == yhorse)
                matrix[i][j] = 0;
            else
                matrix[i][j] += matrix[i][j + 1] + matrix[i + 1][j];

        }
    printf("%lld", matrix[0][0]);
    return 0;
}