此题一开始用暴力做，后来发现斜着走的时候其实暴力不太好写，于是改用搜索写了
1 #include <iostream>
2 #include <stdio.h>
3 #include <memory.h>
4 using namespace std;
5
6 char a[110][110]= {0};
7 int down[110][110]= {0};
8 int up[110][110]= {0};
9
10 int cnt[110][110][4]= {0};
11 int n;
12 int dx[4]= {-1,1,-1, 1};
13 int dy[4]= {-1,1, 1,-1};
14 int dfs(int curx,int cury,int num)
15 {
16 if(curx<0 || cury<0 || curx>=n || cury>=n)
17 return 0;
18 int &ans=cnt[curx][cury][num];
19 if(a[curx][cury]==‘#‘)
20 return ans=0;
21 return ans=1+dfs(curx+dx[num],cury+dy[num],num);
22 }
23
24
25 int getMax()
26 {
27 int ans=0;
28 for(int i=0; i<n; i++)
29 for(int j=0; j<n; j++)
30 {
31 ans=max(ans,down[i][j]+up[i][j]-1);
32 }
33 return ans;
34 }
35
36 int main()
37 {
38 freopen("in.txt","r",stdin);
39
40 while(scanf("%d",&n),n)
41 {
42 memset(down,0,sizeof down);
43 memset(up,0,sizeof up);
44 memset(cnt,0,sizeof cnt);
45 for(int i=0; i<n; i++)
46 scanf("%s",a[i]);
47
48 for(int i=0; i<n; i++)
49 for(int j=0; j<n; j++)
50 {
51 if(a[i][j]==‘#‘)
52 {
53 up[i][j]=down[i][j]=0;
54
55 }
56 else
57 {
58 up[i][j]=down[i][j]=1;
59 if(i!=0)
60 up[i][j]=up[i-1][j]+1;
61 if(j!=0)
62 down[i][j]=down[i][j-1]+1;
63 }
64 }
65
66 int ans=0;
67 ans=max(ans,getMax());
68
69 memset(down,0,sizeof down);
70 memset(up,0,sizeof up);
71 for(int i=n-1; i>=0; i--)
72 for(int j=n-1; j>=0; j--)
73 {
74 if(a[i][j]==‘#‘)
75 {
76 up[i][j]=down[i][j]=0;
77 }
78 else
79 {
80 up[i][j]=down[i][j]=1;
81 if(i!=n-1)
82 up[i][j]=up[i+1][j]+1;
83 if(j!=n-1)
84 down[i][j]=down[i][j+1]+1;
85 }
86 }
87
88 ans=max(ans,getMax());
89
90 //---------------------------
91 memset(down,0,sizeof down);
92 memset(up,0,sizeof up);
93 for(int i=0; i<n; i++)
94 for(int j=n-1; j>=0; j--)
95 {
96 if(a[i][j]==‘#‘)
97 {
98 up[i][j]=down[i][j]=0;
99 }
100 else
101 {
102 up[i][j]=down[i][j]=1;
103 if(i!=0)
104 up[i][j]=up[i-1][j]+1;
105 if(j!=n-1)
106 down[i][j]=down[i][j+1]+1;
107 }
108 }
109
110 ans=max(ans,getMax());
111 //-------------------------------
112 memset(down,0,sizeof down);
113 memset(up,0,sizeof up);
114 for(int i=n-1; i>=0; i--)
115 for(int j=0; j<n; j++)
116 {
117 if(a[i][j]==‘#‘)
118 {
119 up[i][j]=down[i][j]=0;
120 }
121 else
122 {
123 up[i][j]=down[i][j]=1;
124 if(i!=n-1)
125 up[i][j]=up[i+1][j]+1;
126 if(j!=0)
127 down[i][j]=down[i][j-1]+1;
128 }
129 }
130
131 ans=max(ans,getMax());
132 //--------------------------------------------------
133
134 for(int i=0; i<n; i++)
135 for(int j=0; j<n; j++)
136 {
137 for(int num=0; num<4; num++)
138 {
139 dfs(i,j,num);
140 }
141 }
142
143 for(int i=0; i<n; i++)
144 for(int j=0; j<n; j++)
145 {
146 ans=max(ans,cnt[i][j][0]+cnt[i][j][2]-1);
147 ans=max(ans,cnt[i][j][0]+cnt[i][j][3]-1);
148 ans=max(ans,cnt[i][j][1]+cnt[i][j][3]-1);
149 ans=max(ans,cnt[i][j][1]+cnt[i][j][2]-1);
150 }
151
152 cout<<ans<<endl;
153 }
154 return 0;
155 }
hdu5024-Wang Xifeng's Little Plot
时间： 2024-12-29 23:39:24