The Castle

链接

分析：先暴力求出联通块数和最大联通块包含多少，接着对于每个位置判断去掉其上下左右的四个位置的墙之后的最大联通块数，并且记得先选最西，然后选最南的顺序了来输出

  1 /*
  2     PROB:castle
  3     ID:wanghan
  4     LANG:C++
  5 */
  6 #include "iostream"
  7 #include "cstdio"
  8 #include "cstring"
  9 using namespace std;
 10 const int maxn=50+10;
 11 int n,m,a[maxn][maxn],vis[maxn][maxn];
 12 int ans,cnt;
 13 int dx[]={0,-1,0,1};
 14 int dy[]={-1,0,1,0};
 15 int r[]={1,2,4,8};
 16 bool judge(int x,int y){
 17     if(x>n||x<=0||y<=0||y>m)
 18         return false;
 19     return true;
 20 }
 21 void dfs(int x,int y){
 22     if(vis[x][y])
 23         return ;
 24     vis[x][y]=1;
 25     ans++;
 26     for(int i=0;i<4;i++){
 27         int sx=x+dx[i],sy=y+dy[i];
 28         if(judge(sx,sy)){
 29             if((a[x][y]&r[i])==0)
 30                 dfs(sx,sy);
 31         }
 32     }
 33     /*
 34     if((a[x][y]&1)==0)  dfs(x,y-1);
 35     if((a[x][y]&2)==0)  dfs(x-1,y);
 36     if((a[x][y]&4)==0)  dfs(x,y+1);
 37     if((a[x][y]&8)==0)  dfs(x+1,y);*/
 38 }
 39 int main()
 40 {
 41     freopen("castle.in","r",stdin);
 42     freopen("castle.out","w",stdout);
 43     cin>>m>>n;
 44     for(int i=1;i<=n;i++){
 45         for(int j=1;j<=m;j++)
 46             cin>>a[i][j];
 47     }
 48     memset(vis,0,sizeof(vis));
 49     cnt=0;
 50     int res=0;
 51     for(int i=1;i<=n;i++){
 52         for(int j=1;j<=m;j++){
 53             if(!vis[i][j]){
 54                 ++cnt;
 55                 ans=0;
 56                 dfs(i,j);
 57                 res=max(res,ans);
 58             }
 59         }
 60     }
 61     cout<<cnt<<endl;
 62     cout<<res<<endl;
 63     string ch="";
 64     int x,y;
 65     res=0;
 66     for(int j=1;j<=m;j++){
 67         for(int i=n;i>=1;i--){
 68             if((a[i][j]&1)){
 69                 memset(vis,0,sizeof(vis));
 70                 ans=0;
 71                 a[i][j]-=1;
 72                 dfs(i,j);
 73                 if(ans>res){
 74                     res=ans;
 75                     x=i,y=j;
 76                     ch="";
 77                     ch="W";
 78                 }
 79                 a[i][j]+=1;
 80             }
 81             if((a[i][j]&2)){
 82                 memset(vis,0,sizeof(vis));
 83                 ans=0;
 84                 a[i][j]-=2;
 85                 dfs(i,j);
 86                 if(ans>res){
 87                     res=ans;
 88                     x=i,y=j;
 89                     ch="";
 90                     ch="N";
 91                 }
 92                 a[i][j]+=2;
 93             }
 94             if((a[i][j]&4)){
 95                 memset(vis,0,sizeof(vis));
 96                 ans=0;
 97                 a[i][j]-=4;
 98                 dfs(i,j);
 99                 if(ans>res){
100                     res=ans;
101                     x=i,y=j;
102                     ch="";
103                     ch="E";
104                 }
105                 a[i][j]+=4;
106             }
107             if((a[i][j]&8)){
108                 memset(vis,0,sizeof(vis));
109                 ans=0;
110                 a[i][j]-=8;
111                 dfs(i,j);
112                 if(ans>res){
113                     res=ans;
114                     x=i,y=j;
115                     ch="";
116                     ch="S";
117                 }
118                 a[i][j]+=8;
119             }
120         }
121     }
122     cout<<res<<endl;
123     cout<<x<<" "<<y<<" "<<ch<<endl;
124 }
View Code

 

posted @ 2017-07-04 14:41  wolf940509  阅读(128)  评论(0)  编辑  收藏  举报