hihoCoder#1119 小Hi小Ho的惊天大作战：扫雷·二

原题地址

 

没有复杂算法，就是麻烦，写起来细节比较多，比较考验细心，一次AC好开心。

 

代码：

  1 #include <iostream>

  2 #include <vector>

  3 #include <cstring>

  4 #include <cstdlib>

  5 

  6 using namespace std;

  7 

  8 #define SIZE 400

  9 

 10 int N, M;

 11 int map[SIZE][SIZE];

 12 int res[SIZE][SIZE];

 13 

 14 vector<pair<int, int> > unknown_neighbors(int i, int j) {

 15   vector<pair<int, int> > nbs;

 16 

 17   for (int ii = -1; ii < 2; ii++)

 18     for (int jj = -1; jj < 2; jj++) {

 19       int r = ii + i;

 20       int c = jj + j;

 21       if (r >= 0 && r < N && c >= 0 && c < M && (r != i || c != j) && map[r][c] < 0)

 22         nbs.push_back(pair<int, int>(r, c));

 23     }

 24 

 25   return nbs;

 26 }

 27 

 28 vector<pair<int, int> > known_counterparts(int i, int j) {

 29   vector<pair<int, int> > cps;

 30 

 31   for (int ii = -2; ii < 3; ii++) {

 32     for (int jj = -2; jj < 3; jj++) {

 33       int r = ii + i;

 34       int c = jj + j;

 35       if (r >= 0 && r < N && c >= 0 && c < M && (r != i || c != j) && map[r][c] >= 0)

 36         cps.push_back(pair<int, int>(r, c));

 37     }

 38   }

 39 

 40   return cps;

 41 }

 42 

 43 vector<pair<int, int> > differ(vector<pair<int, int> > &a, vector<pair<int, int> > &b) {

 44   vector<pair<int, int> > diff;

 45   int count = 0;

 46 

 47   for (auto pa : a) {

 48     bool found = false;

 49     for (auto pb : b) {

 50       if (pb == pa) {

 51         found = true;

 52         count++;

 53         break;

 54       }

 55     }

 56     if (!found)

 57       diff.push_back(pa);

 58   }

 59 

 60   if (count < b.size())

 61     diff.clear();

 62 

 63   return diff;

 64 }

 65 

 66 void merge_pos(int i, int j, int v) {

 67   if (res[i][j] == -1 || res[i][j] == v)

 68     res[i][j] = v;

 69   else

 70     res[i][j] = -2;

 71 }

 72 

 73 void solve() {

 74   for (int i = 0; i < N; i++)

 75     for (int j = 0; j < M; j++) {

 76       if (map[i][j] == 0) {

 77         vector<pair<int, int> > nbs = unknown_neighbors(i, j);

 78         for (auto p : nbs)

 79           merge_pos(p.first, p.second, 0);

 80       }

 81       else if (map[i][j] > 0) {

 82         vector<pair<int, int> > nbs = unknown_neighbors(i, j);

 83         if (nbs.size() == map[i][j]) {

 84           for (auto p : nbs)

 85             merge_pos(p.first, p.second, 1);

 86           continue;

 87         }

 88 

 89         vector<pair<int, int> > cps = known_counterparts(i, j);

 90         for (auto p : cps) {

 91           vector<pair<int, int> > cp_nbs = unknown_neighbors(p.first, p.second);

 92           if (nbs.size() <= cp_nbs.size() || map[i][j] <= map[p.first][p.second])

 93             continue;

 94           vector<pair<int, int> > diff = differ(nbs, cp_nbs);

 95           if ((int) (nbs.size() - cp_nbs.size()) == map[i][j] - map[p.first][p.second] && !diff.empty())

 96             for (auto dp : diff)

 97               merge_pos(dp.first, dp.second, 1);

 98         }

 99       }

100     }

101 }

102 

103 int main() {

104   int n;

105 

106   cin >> n;

107   while (n--) {

108     cin >> N >> M;

109     for (int i = 0; i < N; i++)

110       for (int j = 0; j < M; j++)

111         cin >> map[i][j];

112     memset(res, -1, SIZE * SIZE * sizeof(int));

113 

114     solve();

115 

116     int mine = 0;

117     int not_mine = 0;

118 

119     for (int i = 0; i < N; i++)

120       for (int j = 0; j < M; j++) {

121         mine += (res[i][j] == 1 ? 1 : 0);

122         not_mine += (res[i][j] == 0 ? 1 : 0);

123       }

124 

125     cout << mine << " " << not_mine << endl;

126   }

127 

128   return 0;

129 }