UVa 439

最新推荐文章于 2021-08-01 10:07:45 发布

climber_studio

最新推荐文章于 2021-08-01 10:07:45 发布

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本文链接：https://blog.youkuaiyun.com/climber_studio/article/details/5812472

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简单bfs，主要是练习bfs架构，以前写usaco本该用bfs的题目都偷懒用了dfs...这是我写得第二道bfs.

#		Problem	Verdict	Language	Run Time	Submission Date
8177314	439	Knight Moves	Accepted	C++	0.028	2010-08-13 12:42:09

[Code]

#include<iostream>
using namespace std;
struct point{int x, y;};
const int MAXN = 100;
struct queue
{
    int f, r, k, n, q[MAXN];
    queue() {memset(q, 0, sizeof(q)); n = k = f= r = 0;}
    void push(int x){q[k++] = x; r++; n++;}
    int pop(){n--; return q[f++];}
    bool empty(){return (f<r) ? 0 : 1;}
};
bool vis[10][10];     point a, b;
int dist[10][10], dx[] = {-1, -2, -2, -1, 1, 2, 2, 1}, dy[] = {2, 1, -1, -2, -2, -1, 1, 2};
void bfs(int x, int y)
{
    queue q;
    int u = x*8 + y;//节点编码
    vis[x][y] = 1;//标记节点已遍历
    dist[x][y] = 0;//初始化距离
    q.push(u);//节点加入队列
    while(!q.empty())
    {
        u = q.pop();//节点出列
        x = u/8; y = u%8;//解码
        for (int d = 0; d < 8; d++)//遍历相关节点
        {
            int nx = x+dx[d], ny = y+dy[d];//利用常量数组表示相关节点
            if (nx>=0 && nx<8 && ny>=0 && ny < 8 && !vis[nx][ny])//判断节点是否符合条件
            {
                int v = nx*8+ny;//编码
                q.push(v);//加入队列
                vis[nx][ny] = 1;//标记节点已遍历
                dist[nx][ny] = dist[x][y]+1;//计算距离
                if (b.x == nx && b.y == ny) return;//判断是否到目标状态
            }
        }
    }
}
int main()
{
    char str[2];
    while(cin.getline(str, 6) != NULL)
    {
        memset(vis, 0, sizeof(vis));
        memset(dist, 0, sizeof(dist));
        a.x = str[0] - 'a';
        a.y = str[1] - '0' - 1;
        b.x = str[3] - 'a';
        b.y = str[4] - '0' - 1;
        bfs(a.x, a.y);   
        printf("To get from %c%d to %c%d takes %d knight moves./n", a.x+'a', a.y+1, b.x+'a', b.y+1, dist[b.x][b.y]);
    }
}