hihoCoder#1054 滑动解锁

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对于每个待枚举的点，检查：

1. 度数检查：是否违反了出度入度限制。因为生成的路径除了首尾节点外，其他节点的出度和入度只能为2

2. 共线检查：是否违反了共线条件。即跨越了尚未枚举过的节点

对于枚举产生的路径，检查：

1. 长度检查：长度是否大于等于4

2. 完整性检查：是否包含了片段中出现的所有边

代码：

#include <iostream>
#include <cstring>

using namespace std;

int impossible[100];

bool integrity_check(int piece[9][9], int path[9][9]) {
  for (int i = 0; i < 9; i++)
    for (int j = 0; j < 9; j++)
        if (piece[i][j] && !path[i][j])
          return false;
  return true;
}

bool degree_check(int piece[9][9], int path[9][9], int b, int e) {
  if (piece[b][e])
    return true;

  int bdegree = 0;
  int edegree = 0;

  for (int i = 0; i < 9; i++) {
    bdegree += path[b][i] || piece[b][i] ? 1 : 0;
    edegree += path[e][i] || piece[e][i] ? 1 : 0;
  }

  return bdegree < 2 && edegree < 2;
}

bool collineation_check(int visited[9], int b, int e) {
  int t = (b + 1) * 10 + e + 1;
  return impossible[t] < 0 || visited[impossible[t] - 1];
}

bool check(int piece[9][9], int path[9][9], int visited[9], int b, int e) {
  if (visited[e])
    return false;

  return degree_check(piece, path, b, e) && collineation_check(visited, b, e);
}

int search(int piece[9][9], int path[9][9], int visited[9], int p, int len) {
  int res = 0;

  if (len >= 4 && integrity_check(piece, path))
    res++;

  for (int i = 0; i < 9 && len < 9; i++) {
    if (check(piece, path, visited, p, i)) {
      visited[i] = 1;
      path[p][i] = path[i][p] = 1;
      res += search(piece, path, visited, i, len + 1);
      path[p][i] = path[i][p] = 0;
      visited[i] = 0;
    }
  }

  return res;
}

int main() {
  int T;
  memset(impossible, -1, sizeof(int) * 100);
  impossible[13] = 2;
  impossible[46] = 5;
  impossible[79] = 8;
  impossible[17]= 4;
  impossible[28]= 5;
  impossible[39]= 6;
  impossible[19]= 5;
  impossible[37]= 5;
  impossible[31]= 2;
  impossible[64]= 5;
  impossible[97]= 8;
  impossible[71]= 4;
  impossible[82]= 5;
  impossible[93]= 6;
  impossible[91]= 5;
  impossible[73]= 5;

  cin >> T;
  while (T--) {
    int N;
    int piece[9][9] = {{0, 0}};
    int path[9][9] = {{0, 0}};
    int visited[9] = {0};
    int res = 0;

    cin >> N;
    for (int i = 0; i < N; i++) {
      int a, b;
      cin >> a >> b;
      piece[a - 1][b - 1] = piece[b - 1][a - 1] = 1;
    }

    for (int i = 0; i < 9; i++) {
      visited[i] = 1;
      res += search(piece, path, visited, i, 1);
      visited[i] = 0;
    }

    cout << res << endl;
  }
  return 0;
}