kickstart2019 round_A B. Parcels

思路：

利用了曼哈頓距離和切比雪夫距離之間的轉化。

參考：

http://www.javashuo.com/article/p-wlfssczx-my.html

http://www.javashuo.com/article/p-clqlbczl-hk.html

之後若是再遇到須要先計算大量曼哈頓距離而後再計算這些曼哈頓距離中的最大值的時候，就要考慮下是否能夠轉化爲切比雪夫距離來優化了。

實現：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef pair<int, int> pii;
 4 const int N = 255;
 5 const int INF = 0x3f3f3f3f;
 6 const int dx[4] = {0, 1, 0, -1};
 7 const int dy[4] = {1, 0, -1, 0};
 8 char a[N][N];
 9 int dis[N][N], vis[N][N];
10 bool check(int k, int r, int c)
11 {
12     memset(dis, 0, sizeof dis);
13     memset(vis, 0, sizeof vis);
14     queue<pii> q;
15     for (int i = 0; i < r; i++)
16     {
17         for (int j = 0; j < c; j++)
18         {
19             if (a[i][j] == '1')
20             {
21                 q.push(pii(i, j));
22                 vis[i][j] = 1;
23                 dis[i][j] = 0;
24             }
25         }
26     }
27     int maxn = 0;
28     while (!q.empty())
29     {
30         pii tmp = q.front(); q.pop();
31         for (int i = 0; i < 4; i++)
32         {
33             int nx = tmp.first + dx[i], ny = tmp.second + dy[i];
34             if (nx >= 0 && nx < r && ny >= 0 && ny < c && !vis[nx][ny])
35             {
36                 vis[nx][ny] = 1;
37                 q.push(pii(nx, ny));
38                 dis[nx][ny] = dis[tmp.first][tmp.second] + 1;
39                 maxn = max(maxn, dis[nx][ny]);
40             }
41         }
42     }
43     if (maxn <= k) return true;
44     int maxx = -INF, maxy = -INF, minx = INF, miny = INF;
45     for (int i = 0; i < r; i++)
46     {
47         for (int j = 0; j < c; j++)
48         {
49             if (dis[i][j] > k)
50             {
51                 maxx = max(maxx, i + j);
52                 minx = min(minx, i + j);
53                 maxy = max(maxy, i - j);
54                 miny = min(miny, i - j);
55             }
56         }
57     }
58     int ans = maxn;
59     for (int i = 0; i < r; i++)
60     {
61         for (int j = 0; j < c; j++)
62         {
63             if (a[i][j] == '1') continue;
64             int tx = i + j, ty = i - j, maxd = -INF;
65             maxd = max(maxd, abs(maxx - tx));
66             maxd = max(maxd, abs(minx - tx));
67             maxd = max(maxd, abs(maxy - ty));
68             maxd = max(maxd, abs(miny - ty));
69             ans = min(ans, maxd);
70         }
71     }
72     return ans <= k;
73 }
74 int main()
75 {
76     int T, r, c;
77     cin >> T;
78     for (int t = 1; t <= T; t++)
79     {
80         cin >> r >> c;
81         for (int i = 0; i < r; i++)
82             for (int j = 0; j < c; j++)
83                 cin >> a[i][j];
84         int L = 0, R = r + c, ans = INF;
85         while (L <= R)
86         {
87             int m = L + R >> 1;
88             if (check(m, r, c))
89             {
90                 R = m - 1;
91                 ans = m;
92             }
93             else L = m + 1;
94         }
95         cout << "Case #" << t << ": " << ans << endl;
96     }
97     return 0;
98 }