USACO 2.1 The Castle(DFS)

感觉代码大体写的还行，就是BUG太多了。。。特别注意输出的优先级，然后了

 1 /*
 2  ID: cuizhe
 3  LANG: C++
 4  TASK: castle
 5  */
 6 #include <cstdio>
 7 #include <cstring>
 8 #include <cmath>
 9 using namespace std;
10 int p[60][61],o[60][60][5],key[61][61],num,n,m;
11 int f[3501];
12 int a[4] = {0,-1,0,1};
13 int b[4] = {-1,0,1,0};
14 void judge(int x,int y,int sum)
15 {
16     int i;
17     for(i = 0; i <= 3; i ++)
18     {
19         if((1<<i)&sum)
20             o[x][y][i] = 1;
21     }
22 }
23 void dfs(int x,int y)
24 {
25     int i;
26     key[x][y] = num;
27     for(i = 0; i <= 3; i ++)
28     {
29         if(x+a[i]<=n&&x+a[i]>=1&&y+b[i]<=m&&y+b[i]>=1&&!key[x+a[i]][y+b[i]]&&!o[x][y][i])
30             dfs(x+a[i],y+b[i]);
31     }
32 }
33 int main()
34 {
35     int i,j,k,t,r,c,d,ans;
36     freopen("castle.in","r",stdin);
37     freopen("castle.out","w",stdout);
38     scanf("%d%d",&m,&n);
39     for(i = 1; i <= n; i ++)
40     {
41         for(j = 1; j <= m; j ++)
42         {
43             scanf("%d",&p[i][j]);
44             judge(i,j,p[i][j]);
45         }
46     }
47     num = 1;
48     for(i = 1; i <= n; i ++)
49     {
50         for(j = 1; j <= m; j ++)
51         {
52             if(!key[i][j])
53             {
54                 dfs(i,j);
55                 num ++;
56             }
57         }
58     }
59     printf("%d\n",num-1);
60     for(i = 1; i <= n; i ++)
61     {
62         for(j = 1; j <= m; j ++)
63         {
64             f[key[i][j]]++;
65         }
66     }
67     t = 1;
68     for(i = 1; i <= num-1; i ++)
69     {
70         if(t < f[i])
71             t = f[i];
72     }
73     printf("%d\n",t);
74     ans = t;
75     for(j = 1; j <= m; j ++)//优先级
76     {
77         for(i = n; i >= 1; i --)
78         {
79             for(k = 1; k <= 2; k ++)
80             {
81                 if(i+a[k]<=n&&i+a[k]>=1&&j+b[k]>=1&&j+b[k]<=m&&key[i][j]!=key[i+a[k]][j+b[k]])
82                 {
83                     if(ans < f[key[i][j]]+f[key[i+a[k]][j+b[k]]])
84                     {
85                         ans = f[key[i][j]]+f[key[i+a[k]][j+b[k]]];
86                         r = i;
87                         c = j;
88                         d = k;
89                     }
90                 }
91             }
92         }
93     }
94     printf("%d\n",ans);
95     printf("%d %d ",r,c);
96     if(d == 1) printf("N\n");
97     else printf("E\n");
98     return 0;
99 }

 

犯各种很2的错误。。

posted @ 2012-11-05 11:50  Naix_x  阅读(126)  评论(0)  编辑  收藏  举报