CF1105D Kilani and the Game

思路：

模拟多源bfs。

实现：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef pair<int, int> pii;
 4 char a[1005][1005];
 5 const int dx[4] = {1, 0, -1, 0};
 6 const int dy[4] = {0, 1, 0, -1};
 7 int s[10], ans[10], l[10], d[1005][1005];
 8 inline bool check(int x, int y, int n, int m)
 9 {
10     return x >= 1 && x <= n && y >= 1 && y <= m && a[x][y] == ‘.‘;
11 }
12 int main()
13 {
14     int n, m, p;
15     while (cin >> n >> m >> p)
16     {
17         memset(ans, 0, sizeof ans);
18         memset(d, 0, sizeof d);
19         memset(l, 0, sizeof l);
20         for (int i = 0; i < p; i++) cin >> s[i];
21         vector<queue<pii>> v(p);
22         for (int i = 1; i <= n; i++)
23         {
24             for (int j = 1; j <= m; j++)
25             {
26                 cin >> a[i][j];
27                 if (a[i][j] >= ‘1‘ && a[i][j] <= ‘9‘)
28                 {
29                     v[a[i][j] - ‘1‘].push(make_pair(i, j));
30                 }
31             }
32         }
33         while (true)
34         {
35             bool flg = true;
36             for (int i = 0; i < p; i++)
37             {
38                 while (!v[i].empty())
39                 {
40                     pii t = v[i].front();
41                     int x = t.first, y = t.second;
42                     if (d[x][y] >= l[i] + s[i]) break;
43                     v[i].pop();
44                     for (int k = 0; k < 4; k++)
45                     {
46                         int nx = x + dx[k], ny = y + dy[k];
47                         if (check(nx, ny, n, m))
48                         {
49                             v[i].push(make_pair(nx, ny));
50                             a[nx][ny] = char(‘1‘ + i);
51                             d[nx][ny] = d[x][y] + 1;
52                         }
53                     }
54                 }
55                 l[i] += s[i];
56                 if (v[i].size() > 0) flg = false;
57             }
58             if (flg) break;
59         }
60         for (int i = 1; i <= n; i++)
61         {
62             for (int j = 1; j <= m; j++)
63             {
64                 if (a[i][j] >= ‘1‘ && a[i][j] <= ‘9‘) ans[a[i][j] - ‘0‘]++;
65             }
66         }
67         for (int i = 1; i <= p; i++) cout << ans[i] << " ";
68         cout << endl;
69     }
70     return 0;
71 }

原文地址：https://www.cnblogs.com/wangyiming/p/10848604.html