计蒜客 11064 大钉骑马走江湖

最新推荐文章于 2020-04-20 00:38:12 发布

原创最新推荐文章于 2020-04-20 00:38:12 发布 · 469 阅读

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CC 4.0 BY-SA版权

本文介绍了一种利用广度优先搜索(BFS)算法解决特殊移动限制下的图中两点间最短路径问题的方法。该问题设定在一个允许类似马棋子跳跃的网格上，寻找从起点到终点的最短步数。

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原题

Description

一个图，只能像马那么跳，求起点到终点最短时间

Algorithm

广搜

Ps

一开始写错一个符合狂WA

Code

#include <cstring>
#include <cstdio>
#include <iostream>
#include <queue>
using namespace std;
int n, m;
const int maxn = 100 + 9;
const int inf = 1e8;
char g[maxn][maxn];
struct V
{
  int x, y;
  V(int x, int y) : x(x), y(y){}
  V(){}
};
int a[maxn][maxn];
void print()
{
  for (int i = 0; i < n; i++)
  {
    cout << endl;
    for (int j = 0; j < m; j++)
      if (a[i][j] == inf) cout << '0'; else cout << a[i][j];
  }
  cout << endl;
}
bool no(const int &x, const int &y)
{
  if ((x < 0) ||
      (x >= n)||
      (y < 0) ||
      (y >= m)||
      (g[x][y] == '#')) return false;
  return true;
}
void solve()
{
  memset(g, 0, sizeof(g));
  memset(a, 0, sizeof(a));
  for (int i = 0; i < n; i++) scanf("%s", g[i]);
  queue<V> v;
  V start, ee;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
    {
      a[i][j] = inf;
      if (g[i][j] == 's')
      {
        start.x = i;
        start.y = j;
      }
      if (g[i][j] == 'e')
      {
        ee.x = i;
        ee.y = j;
      }
    }
  v.push(start);
  a[start.x][start.y] = 0;
  while (!v.empty())
  {
    V now = v.front();
    v.pop();
    int x = now.x, y = now.y;
    int s = a[x][y] + 1;
    if (no(x - 2, y + 1) && no(x - 1, y) && a[x - 2][y + 1] > s)
    {
      a[x - 2][y + 1] = s;
      v.push(V(x - 2, y + 1));
    }
    if (no(x - 2, y - 1) && no(x - 1, y) && a[x - 2][y - 1] > s)
    {
      a[x - 2][y - 1] = s;
      v.push(V(x - 2, y - 1));
    }
    if (no(x - 1, y + 2) && no(x, y + 1) && a[x - 1][y + 2] > s)
    {
      a[x - 1][y + 2] = s;
      v.push(V(x - 1, y + 2));
    }
    if (no(x + 1, y + 2) && no(x, y + 1) && a[x + 1][y + 2] > s)
    {
      a[x + 1][y + 2] = s;
      v.push(V(x + 1, y + 2));
    }
    if (no(x + 2, y + 1) && no(x + 1, y) && a[x + 2][y + 1] > s)
    {
      a[x + 2][y + 1] = s;
      v.push(V(x + 2, y + 1));
    }
    if (no(x + 2, y - 1) && no(x + 1, y) && a[x + 2][y - 1] > s)
    {
      a[x + 2][y - 1] = s;
      v.push(V(x + 2, y - 1));
    }
    if (no(x + 1, y - 2) && no(x, y - 1) && a[x + 1][y - 2] > s)
    {
      a[x + 1][y - 2] = s;
      v.push(V(x + 1, y - 2));
    }
    if (no(x - 1, y - 2) && no(x, y - 1) && a[x - 1][y - 2] > s)
    {
      a[x - 1][y - 2] = s;
      v.push(V(x - 1, y - 2));
    }
  }
  if (a[ee.x][ee.y] == inf) a[ee.x][ee.y] = -1;
  printf("%d\n", a[ee.x][ee.y]);
}
int main()
{
//  freopen("input.txt", "r", stdin);
  while (scanf("%d%d\n", &n, &m) != EOF)
    solve();
}