CoderForces 327D Block Tower

Portal:http://codeforces.com/problemset/problem/327/D

一座红塔200人，一座蓝塔100人，只有与蓝塔相邻才可以建红塔。

'.'处可建塔 '#'处不可建塔

可以随便毁塔

在建造能够容纳最多人数的塔集合情况下，输出对塔的操作（SPJ）

这是一道水题

这是一道卡输入输出的题

DFS就好，反正就是在联通分量里炸割顶留根

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<set>
 4 #include<cstdio>
 5 #include<cstdlib>
 6 #include<cmath>
 7 #include<cstring>
 8 #include<vector>
 9 using namespace std;
10 #define FOR(i,j,k) for(int i=j;i<=k;i++)
11 #define FORD(i,j,k) for(int i=j;i>=k;i--)
12 #define LL long long
13 #define SZ(x) int(x.size())
14 #define maxm 510
15 #define maxn 510
16 string x;
17 int dx[]={1,0,-1,0};
18 int dy[]={0,1,0,-1};
19 int m,n,sum;
20 vector<char> al;
21 vector<int> ax,ay;
22 char ii[maxm][maxn];
23 int a[maxm][maxn],vis[maxm][maxn],col[maxm][maxn];
24 bool cr(int xx,int yy)
25 {
26     FOR(i,0,3)
27     if(xx+dx[i]>0&&xx+dx[i]<=m&&yy+dy[i]>0&&yy+dy[i]<=n)
28     if(col[xx+dx[i]][yy+dy[i]]==1) return true;
29     return false;
30 }
31 void dfs(int x,int y)
32 {
33     vis[x][y]=1;
34     /*if(cr(x,y)) {al.push_back('R');ax.push_back(x);ay.push_back(y);col[x][y]=2;sum++;}
35     else */{al.push_back('B');ax.push_back(x);ay.push_back(y);col[x][y]=1;sum++;}
36     FOR(i,0,3)
37     if(x+dx[i]>0&&x+dx[i]<=m&&y+dy[i]>0&&y+dy[i]<=n)
38     if(a[x+dx[i]][y+dy[i]]) if(!vis[x+dx[i]][y+dy[i]]) dfs(x+dx[i],y+dy[i]);
39     if(col[x][y]==1) if(cr(x,y)) { al.push_back('D');ax.push_back(x);ay.push_back(y);
40      al.push_back('R');ax.push_back(x);ay.push_back(y);col[x][y]=2;sum+=2;}
41 }
42 int main()
43 {
44 cin>>m>>n;
45 FOR(i,1,m)
46 {scanf("%s",&ii[i][1]);
47 FOR(j,1,n)
48 {
49 a[i][j]=(ii[i][j]=='.');
50 }
51 }
52 FOR(i,1,m)
53 FOR(j,1,n)
54 if(a[i][j]&&!vis[i][j]) dfs(i,j);
55 cout<<sum<<endl;
56 FOR(i,0,SZ(al)-1)
57 printf("%c %d %d
",al[i],ax[i],ay[i]);
58 return 0;
59 }

特别丑的原代码（AC）

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<set>
 4 #include<cstdio>
 5 #include<cstdlib>
 6 #include<cmath>
 7 #include<cstring>
 8 #include<sstream>
 9 using namespace std;
10 #define FOR(i,j,k) for(int i=j;i<=k;i++)
11 #define FORD(i,j,k) for(int i=j;i>=k;i--)
12 #define LL long long
13 #define SZ(x) int(x.size())
14 #define maxm 510
15 #define maxn 510
16 char x;
17 stringstream ans;
18 int dx[]={1,0,-1,0};
19 int dy[]={0,1,0,-1};
20 int m,n,sum;
21 char ii[maxm][maxn];
22 int a[maxm][maxn],vis[maxm][maxn],col[maxm][maxn];
23 bool cr(int xx,int yy)
24 {
25     FOR(i,0,3)
26     if(xx+dx[i]>0&&xx+dx[i]<=m&&yy+dy[i]>0&&yy+dy[i]<=n)
27     if(col[xx+dx[i]][yy+dy[i]]==1) return true;
28     return false;
29 }
30 void dfs(int x,int y)
31 {
32     vis[x][y]=1;
33     ans<<"B "<<x<<' '<<y<<"
";col[x][y]=1;sum++;
34     FOR(i,0,3)
35     if(x+dx[i]>0&&x+dx[i]<=m&&y+dy[i]>0&&y+dy[i]<=n)
36     if(a[x+dx[i]][y+dy[i]]) if(!vis[x+dx[i]][y+dy[i]]) dfs(x+dx[i],y+dy[i]);
37     if(col[x][y]==1) if(cr(x,y)) {     ans<<"D "<<x<<' '<<y<<"
";
38      ans<<"R "<<x<<' '<<y<<"
";col[x][y]=2;sum+=2;}
39 }
40 int main()
41 {
42 cin>>m>>n;
43 FOR(i,1,m)
44 FOR(j,1,n)
45 {
46     cin>>x;
47     a[i][j]=(x=='.');
48 }
49 FOR(i,1,m)
50 FOR(j,1,n)
51 if(a[i][j]&&!vis[i][j]) dfs(i,j);
52 cout<<sum<<endl;
53 cout<<ans.str();
54 return 0;
55 }

改后依然很丑的代码

输出黑科技sstream orz

#include<sstream> x;
x<<.....;
cout<<x.str();

输入黑科技

scanf("%s",&string[i][1]);