acwing-195. 骑士精神

发表于 分类于 阅读次数： Waline： 本文字数： 622 阅读时长 ≈ 2 分钟

考点

  • 双向BFS
  • IDAstar

题解

见思路

思路

每次其实移动的是空白格，不是棋子

IDAstar

题目告诉你最多15步，考虑IDAstar

由于每次最多把一个棋子移到合适的地方，所以估价函数就等于当前未到目标位置棋子的数量

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#include <bits/stdc++.h>
using namespace std;
const int maxn = 5;
const int dx[8] = {-2, -1, 1, 2, 2, 1, -1, -2};
const int dy[8] = {1, 2, 2, 1, -1, -2, -2, -1};
const char ans[maxn][maxn + 1] = {"11111", "01111", "00*11", "00001", "00000"};
char g[maxn][maxn];

int f() {
  int res = 0;
  for (int i = 0; i < 5; ++i) {
    for (int j = 0; j < 5; ++j) {
      if (g[i][j] != '*' && g[i][j] != ans[i][j]) ++res;
    }
  }
  return res;
}

bool ida(int x, int y, int depth, int mxdepth) {
  if (depth + f() > mxdepth) return false;
  if (!f()) return true;
  for (int i = 0; i < 8; ++i) {
    int nx = x + dx[i], ny = y + dy[i];
    if (nx < 0 || ny < 0 || nx > 4 || ny > 4) continue;
    swap(g[x][y], g[nx][ny]);
    if (ida(nx, ny, depth + 1, mxdepth)) return true;
    swap(g[x][y], g[nx][ny]);
  }
  return false;
}

int main() {
  int t, x, y;
  cin >> t;
  while (t--) {
    for (int i = 0; i < 5; ++i) {
      for (int j = 0; j < 5; ++j) {
        cin >> g[i][j];
        if (g[i][j] == '*') x = i, y = j;
      }
    }
    int depth = 0;
    while (depth < 16 && !ida(x, y, 0, depth)) ++depth;
    cout << (depth < 16 ? depth : -1) << endl;
  }
  return 0;
}

双向BFS

由于已知起始状态和目标状态，且没有什么有效剪枝，可以考虑双向BFS，直接套模板即可

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#include <bits/stdc++.h>
using namespace std;
const int dx[8] = {-2, -1, 1, 2, 2, 1, -1, -2};
const int dy[8] = {1, 2, 2, 1, -1, -2, -2, -1};
string a, b = "1111101111002110000100000";

struct node {
  string state_;
  int x_, y_, step_;
  node(string state = "", int x = 0, int y = 0, int step = 0)
      : state_(state), x_(x), y_(y), step_(step) {}
};

bool valid(int x, int y) { return x >= 0 && y >= 0 && x < 5 && y < 5; }

int extend(queue<node> &aq, queue<node> &bq, unordered_map<string, bool> &av,
           unordered_map<string, bool> &bv) {
  int sz = aq.size();
  while (sz--) {
    string s = aq.front().state_;
    int x = aq.front().x_, y = aq.front().y_, step = aq.front().step_;
    aq.pop();
    for (int i = 0; i < 8; ++i) {
      int nx = x + dx[i], ny = y + dy[i];
      if (!valid(nx, ny)) continue;
      string ns = s;
      swap(ns[x * 5 + y], ns[nx * 5 + ny]);
      if (av.count(ns)) continue;
      if (bv.count(ns)) return bq.front().step_ + step + 1;
      av[ns] = 1;
      aq.emplace(ns, nx, ny, step + 1);
    }
  }
  return -1;
}

int bfs(int ax, int ay) {
  unordered_map<string, bool> av, bv;
  queue<node> aq, bq;
  if (a == b) return 0;
  av[a] = 1, bv[b] = 1;
  aq.push({a, ax, ay, 0}), bq.push({b, 2, 2, 0});
  int res = 0, cnt = 1;
  while (aq.size() && bq.size()) {
    if (aq.size() < bq.size()) {
      res = extend(aq, bq, av, bv);
    } else {
      res = extend(bq, aq, bv, av);
    }
    if (~res) return res;
    if (++cnt > 15) break;
  }
  return -1;
}

int main() {
  int t, x, y, res;
  char c;
  cin >> t;
  while (t--) {
    a = "";
    for (int i = 0; i < 5; ++i) {
      for (int j = 0; j < 5; ++j) {
        cin >> c;
        if (c == '*') c = '2', x = i, y = j;
        a += c;
      }
    }
    res = bfs(x, y);
    cout << res << endl;
  }
  return 0;
}