POJ 2243-Knight Moves

http://poj.org/problem?id=2243

我是在看staginner大牛的博客的时候看到这道题的，因为看到了BFS，所以就拿来做了，但是发现

好像之前没写过BFS这玩意，所以就基本照着搬了一遍他的代码，自己写了一下，理解了下队列和广搜。

题目要我们找到从一个点到另一个点的骑士移动的步数，按照staginner的做法是记录在找到终点之前

的所有点到起点的步数。

#include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;

char a[5], b[5];
int x1, y1, x2, y2;
int dist[10][10], qx[100], qy[100];
int dx[] = { -1, -2, -2, -1, 1, 2, 2, 1};
int dy[] = { -2, -1, 1, 2, 2, -1, 1, -2};

int main()
{
int x, y, nx, ny, front, rear;
while( cin >> a >> b)
    {
        x1 = a[0] - 'a';
        y1 = a[1] - '1';
        x2 = b[0] - 'a';
        y2 = b[1] - '1';
        memset( dist, -1, sizeof( dist) );
        dist[x1][y1] = 0;
        front = rear = 0;
        qx[rear] = x1;
        qy[rear] = y1;
        rear ++;
while( front < rear)
        {
            x = qx[front];
            y = qy[front];
if( x == x2 && y == y2)
break;
            front ++;
for( int i = 0; i < 8; i ++)
            {
                nx = x + dx[i];
                ny = y + dy[i];
if( dist[nx][ny] < 0 && nx >= 0 && nx < 8 && ny >= 0 && ny < 8)
                {
                    dist[nx][ny] = dist[x][y] + 1;
                    qx[rear] = nx;
                    qy[rear] = ny;
                    rear ++;
                }
            }
        }
int steps = dist[x2][y2];
        printf( "To get from %s to %s takes %d knight moves.\n", a, b, steps);
    }
return 0;
}

跟着集训手册做，又做到了这个题：

/*Accepted    176K    188MS    C++    1287B    2012-07-23 13:27:47*/
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<queue>
#include<iostream>
using namespace std;
const int MAXN = 10;
const int dx[] = { 1, 1, -1, -1, 2, 2, -2, -2};
const int dy[] = { 2, -2, 2, -2, 1, -1, 1, -1};
typedef pair<int, int> pii;

int d[MAXN][MAXN];
char a[5], b[5];
int x1, y1, x2, y2;

void bfs()
{
    int i, x, y, nx, ny;
    queue<pii> q;
    pii u;
    d[x1][y1] = 0;
    u.first = x1, u.second = y1;
    q.push(u);
    while(!q.empty())
    {
        u = q.front(); q.pop();
        x = u.first, y = u.second;
        if( x == x2 && y == y2) break;
        for( i = 0; i < 8; i ++)
        {
            nx = x + dx[i];
            ny = y + dy[i];
            if( nx <= 8 && nx > 0 && ny <= 8 && ny > 0 && d[nx][ny] < 0)
            {
                d[nx][ny] = d[x][y] + 1;
                u.first = nx, u.second = ny;
                q.push(u);
            }
        }
    }
}

int main()
{
    while( scanf( "%s%s", a, b) == 2)
    {
        x1 = a[1] - '0';
        y1 = a[0] - 'a' + 1;
        x2 = b[1] - '0';
        y2 = b[0] - 'a' + 1;
        memset( d, -1, sizeof d);
        bfs();
        printf( "To get from %s to %s takes %d knight moves.\n", a, b, d[x2][y2]);
    }
    return 0;
}

 

 

转载于:https://www.cnblogs.com/Yu2012/archive/2011/11/14/2249008.html