pku1915

最新推荐文章于 2025-08-08 11:34:53 发布

henry11

最新推荐文章于 2025-08-08 11:34:53 发布

阅读量421

点赞数

CC 4.0 BY-SA版权

文章标签：数据结构 null delete user

本文链接：https://blog.youkuaiyun.com/henry11/article/details/4443661

本文通过对比链表和数组实现的BFS算法在骑士走法问题上的表现，展示了数据结构选择的重要性。使用数组作为队列的BFS算法不仅通过了测试，而且运行效率更高。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Source Code

Problem: 1915		User: henry11
Memory: 1768K		Time: 94MS
Language: G++		Result: Accepted

Source Code

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

struct node
{
    int a, b;
    int next;
}; 
node queue[100000];
int head, deap, tail;
int direct[8][2]={-2, 1, -1, 2, 1, 2, 2, 1,
                   2, -1, 1, -2,-1, -2, -2, -1};    
int chessBoard[301][301];
int count, length;

void initQueue(int x, int y)
{
    queue[0].a = x;
    queue[0].b = y;
    queue[0].next = -1;
    head = deap = tail = 0;
}    

void insertNode(int x, int y)
{
    tail++;
    queue[tail].a = x;
    queue[tail].b = y;
    chessBoard[x][y] = 1;
    queue[tail].next = -1;
}    

int BFS(int sa, int sb, int da, int db)
{
    int n;
    int i, ta = sa, tb = sb;
    initQueue(ta, tb);
    count = 0;
    if(ta == da && tb == db)return count;
    while(head != -1)
    {
        for(i=0; i<8; i++)
        {
            ta = direct[i][0]+queue[head].a;
            tb = direct[i][1]+queue[head].b;
            //printf("firsttest  %d %d/n", ta, tb);
            if(ta>=0 && ta<length && tb>=0 && tb<length)
            {
                if(ta == da && tb == db)return count+1;
                if(!chessBoard[ta][tb])insertNode(ta, tb);
                //printf("secondtest  %d/n", chessBoard[ta][tb]);
            } 
            //scanf("%d", &n);   
        }    
        if(deap == head)
        {
            count++;
            deap = tail;
        }    
        head++;
        //printf("thirdtest  %d/n", count);
    }    
}    

int main()
{
    int n, s1, s2, d1, d2;
    scanf("%d", &n);
    while(n--)
    {
        memset(chessBoard, 0, sizeof(chessBoard));
        scanf("%d", &length);
        scanf("%d%d%d%d", &s1, &s2, &d1, &d2);
        printf("%d/n", BFS(s1, s2, d1, d2));
        //scanf("%d", &length);
    }
    return 1;
}

应该是很简单的BFS，一开始用链表写超时了用数组做队列就过了。看来以后对数据结构的选择要慎之又慎。差距很大。

还有他们讲用双向DFS，知道是什么，但从来没写过这个，需要多练练啊，不知道还有没有什么更好的办法。

下面是用链表写的，超时，但结果是对的，也算是一个经验。

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

struct node
{
int a, b;
node *next;
}*queue, *deap, *tail;

int direct[8][2]={-2, 1, -1, 2, 1, 2, 2, 1,
2, -1, 1, -2,-1, -2, -2, -1};

int chessBoard[301][301];
int count, length;

void initQueue(int x, int y)
{
node *p;
p = new node;
p->a = x;
p->b = y;
chessBoard[x][y] = 1;
p->next = NULL;
queue = deap = tail = p;
}

void insertNode(int x, int y)
{
node *p;
p = new node;
p->a = x;
p->b = y;
p->next = NULL;
chessBoard[x][y] = 1;
tail->next = p;
tail = p;
}

void deleteNode()
{
node *p;
p = queue;
queue = p->next;
delete p;
}

int BFS(int sa, int sb, int da, int db)
{
int n;
int ta = sa, tb = sb;
int i, j;
initQueue(ta, tb);
//printf("%d %d/n", queue->a, queue->b);
count = 0;
if(ta == da && tb == db)
{
return count;
}
while(queue)
{
for(i=0; i<8; i++)
{
ta = direct[i][0]+queue->a;
tb = direct[i][1]+queue->b;
//printf("firsttest %d %d %d/n", i, ta, tb);
if(ta>=0 && ta<length && tb>=0 && tb<length)
{
if(ta == da && tb == db)
{
return count+1;
}
//printf("secondtest %d/n", chessBoard[ta][tb]);
if(!chessBoard[ta][tb])
{
insertNode(ta, tb);
}
}
//printf("thirdtest %d %d/n", tail->a, tail->b);
//scanf("%d", &n);
}
if(deap == queue)
{
count++;
deap = tail;
}
deleteNode();
//printf("%d/n", count);
}
}

int main()
{
//int i, j;
double start, complete;
start = (double)clock();
int n, s1, s2, d1, d2;
scanf("%d", &n);
while(n--)
{
memset(chessBoard, 0, sizeof(chessBoard));
scanf("%d", &length);
scanf("%d%d%d%d", &s1, &s2, &d1, &d2);
printf("%d/n", BFS(s1, s2, d1, d2));
//scanf("%d", &length);
}
complete = (double)clock();
printf("%.4fms/n", start-complete);
scanf("%d", &n);
return 1;
}