題意：人在N，牛在K，人要去找牛，求最短路徑。人在X，可以前往X+1X−12×X。

思路：裸BFS，直接搜即可。

程式碼：

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <math.h>
#include <queue>

using namespace std;

const int N = 2e5 + 10;
const int INF = 0x3f3f3f3f;

typedef pair<int
 
, int> P;

int dis[N];

int bfs(int s, int e) {
  queue<P> q;
  q.push(make_pair(s, 0));
  while (!q.empty()) {
    P u = q.front();
    q.pop();
    if (u.first == e) {
      return u.second;
    }
    for (int i = -1; i <= 1; i += 2) {
      P t;
      t.first = u.first + i;
      t.second = u.second + 1
 
;
      if (t.first >= 0 && t.first <= 2 * e && dis[t.first] > t.second) {
        q.push(t);
        dis[t.first] = t.second;
      }
    }
    P v;
    v.first = u.first * 2;
    v.second = u.second + 1;
    if (v.first >= 0 && v.first <= 2 * e && dis[v.first] > v.second) {
      q.push(v);
      dis[v.first] = v.second;
    }
  }
  return
 
 -1;
}

int main() {
  int n, k;
  while (scanf("%d%d", &n, &k) != EOF) {
    memset(dis, INF, sizeof(dis));
    if (n >= k)
      printf("%d\n", n - k);
    else
      printf("%d\n", bfs(n, k));
  }
  return 0;
}

題意：三維空間，從S到T的最短路。

思路：裸BFS，直接搜。

程式碼：

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <math.h>
#include <queue>

using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 5e1 + 10;

struct Node {
  int x, y, z;
  int step;
};

int x[4] = {0, 1, 0, -1};
int y[4] = {1, 0, -1, 0};

int dis[N][N][N];
char _map[N][N][N];
int l, r, c;

bool inFiled(Node now) {
  if (now.x >= 0 && now.x < r && now.y >= 0 
    && now.y < c && now.z >= 0 && now.z < l)
    return true;
  return false;
}

int bfs(Node s) {
  queue<Node> q;
  q.push(s);
  dis[s.x][s.y][s.z] = s.step;
  while (!q.empty()) {
    Node now = q.front();
    q.pop();
    if (_map[now.z][now.x][now.y] == 'E')
      return now.step;
    // cout << "now " << now.x << " " << now.y << " " << now.z << endl;
    Node to;
    for (int j = 0; j < 4; j++) {
      to.x = now.x + x[j];
      to.y = now.y + y[j];
      to.z = now.z;
      to.step = now.step + 1;
      if (inFiled(to) && _map[to.z][to.x][to.y] != '#' 
        && dis[to.z][to.x][to.y] > to.step) {
        dis[to.z][to.x][to.y] = to.step;
        q.push(to);
      }
    }
    for (int i = -1; i <= 1; i++) {
      to.x = now.x;
      to.y = now.y;
      to.z = now.z + i;
      to.step = now.step + 1;
      if (inFiled(to) && _map[to.z][to.x][to.y] != '#'
        && dis[to.z][to.x][to.y] > to.step) {
        dis[to.z][to.x][to.y] = to.step;
        q.push(to);
      }
    }
  }
  return -1;
}

int main() {
  while (scanf("%d%d%d", &l, &r, &c) != EOF) {
    if (!l && !r && !c)
      break;
    memset(dis, INF, sizeof(dis));
    for (int i = 0; i < l; i++) 
      for (int j = 0; j < r; j++)
        scanf("%s", _map[i][j]);
    Node st;
    for (int i = 0; i < l; i++) {
      for (int j = 0; j < r; j++) {
        for (int k = 0; k < c; k++) {
          if (_map[i][j][k] == 'S') {
            st.x = j, st.y = k, st.z = i;
            st.step = 0;
          }
        }
      }
    }
    int ans = bfs(st);
    if (ans == -1)
      printf("Trapped!\n");
    else
      printf("Escaped in %d minute(s).\n", ans);
  }
  return 0;
}