bzoj2252 矩阵距离

很好奇这种 普及- 的题为什么会是权限题......

我一开始想用枚举 + 搜索，看书后发现自己脑抽了。直接BFS即可。

 1 #include <cstdio>
 2 #include <queue>
 3 #include <cstring>
 4 
 5 struct Sta {
 6     int x, y, step;
 7     Sta(int x = 0, int y = 0) {
 8         this->x = x;
 9         this->y = y;
10     }
11 };
12 
13 const int N = 1010;
14 int G[N][N], dis[N][N];
15 
16 const int dx[4] = {0, 0, 1, -1};
17 const int dy[4] = {1, -1, 0, 0};
18 
19 int main() {
20     int m, n;
21     std::queue<Sta> Q;
22     memset(dis, -1, sizeof(dis));
23     scanf("%d%d", &n, &m);
24     for(int i = 1; i <= n; i++) {
25         for(int j = 1; j <= m; j++) {
26             char c = getchar();
27             while(c < '0' || c > '1') {
28                 c = getchar();
29             }
30             G[i][j] = c - 48;
31             if(G[i][j]) {
32                 dis[i][j] = 0;
33                 Q.push(Sta(i, j));
34             }
35         }
36     }
37 
38     for(int i = 0; i <= n + 1; i++) {
39         dis[i][0] = dis[i][m + 1] = 0;
40     }
41     for(int i = 1; i <= m; i++) {
42         dis[0][i] = dis[n + 1][i] = 0;
43     }
44 
45     while(!Q.empty()) {
46         Sta op = Q.front();
47         Q.pop();
48         for(int i = 0; i <= 3; i++) {
49             int tx = op.x + dx[i];
50             int ty = op.y + dy[i];
51             if(dis[tx][ty] >= 0) {
52                 continue;
53             }
54             dis[tx][ty] = dis[op.x][op.y] + 1;
55             Q.push(Sta(tx, ty));
56         }
57     }
58 
59     for(int i = 1; i <= n; i++) {
60         for(int j = 1; j <= m; j++) {
61             printf("%d ", dis[i][j]);
62         }
63         if(i < n) {
64             puts("");
65         }
66     }
67 
68     return 0;
69 }
AC代码

代码里的G[N][N]应该是多余的...

posted @ 2018-06-19 18:00  garage  阅读(120)  评论(0)  编辑  收藏  举报