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

 #include<iostream>
 #include<algorithm>
 #include<set>
 #include<cstdio>
 #include<cstdlib>
 #include<cmath>
 #include<cstring>
 #include<vector>
 using namespace std;
 #define FOR(i,j,k) for(int i=j;i<=k;i++)
 #define FORD(i,j,k) for(int i=j;i>=k;i--)
 #define LL long long
 #define SZ(x) int(x.size())
 #define maxm 510
 #define maxn 510
 string x;
 int dx[]={,,-,};
 int dy[]={,,,-};
 int m,n,sum;
 vector<char> al;
 vector<int> ax,ay;
 char ii[maxm][maxn];
 int a[maxm][maxn],vis[maxm][maxn],col[maxm][maxn];
 bool cr(int xx,int yy)
 {
     FOR(i,,)
     if(xx+dx[i]>&&xx+dx[i]<=m&&yy+dy[i]>&&yy+dy[i]<=n)
     if(col[xx+dx[i]][yy+dy[i]]==) return true;
     return false;
 }
 void dfs(int x,int y)
 {
     vis[x][y]=;
     /*if(cr(x,y)) {al.push_back('R');ax.push_back(x);ay.push_back(y);col[x][y]=2;sum++;}
     else */{al.push_back('B');ax.push_back(x);ay.push_back(y);col[x][y]=;sum++;}
     FOR(i,,)
     if(x+dx[i]>&&x+dx[i]<=m&&y+dy[i]>&&y+dy[i]<=n)
     if(a[x+dx[i]][y+dy[i]]) if(!vis[x+dx[i]][y+dy[i]]) dfs(x+dx[i],y+dy[i]);
     if(col[x][y]==) if(cr(x,y)) { al.push_back('D');ax.push_back(x);ay.push_back(y);
      al.push_back('R');ax.push_back(x);ay.push_back(y);col[x][y]=;sum+=;}
 }
 int main()
 {
 cin>>m>>n;
 FOR(i,,m)
 {scanf("%s",&ii[i][]);
 FOR(j,,n)
 {
 a[i][j]=(ii[i][j]=='.');
 }
 }
 FOR(i,,m)
 FOR(j,,n)
 if(a[i][j]&&!vis[i][j]) dfs(i,j);
 cout<<sum<<endl;
 FOR(i,,SZ(al)-)
 printf("%c %d %d\n",al[i],ax[i],ay[i]);
 return ;
 }

特别丑的原代码（AC）

 #include<iostream>
 #include<algorithm>
 #include<set>
 #include<cstdio>
 #include<cstdlib>
 #include<cmath>
 #include<cstring>
 #include<sstream>
 using namespace std;
 #define FOR(i,j,k) for(int i=j;i<=k;i++)
 #define FORD(i,j,k) for(int i=j;i>=k;i--)
 #define LL long long
 #define SZ(x) int(x.size())
 #define maxm 510
 #define maxn 510
 char x;
 stringstream ans;
 int dx[]={,,-,};
 int dy[]={,,,-};
 int m,n,sum;
 char ii[maxm][maxn];
 int a[maxm][maxn],vis[maxm][maxn],col[maxm][maxn];
 bool cr(int xx,int yy)
 {
     FOR(i,,)
     if(xx+dx[i]>&&xx+dx[i]<=m&&yy+dy[i]>&&yy+dy[i]<=n)
     if(col[xx+dx[i]][yy+dy[i]]==) return true;
     return false;
 }
 void dfs(int x,int y)
 {
     vis[x][y]=;
     ans<<"B "<<x<<' '<<y<<"\n";col[x][y]=;sum++;
     FOR(i,,)
     if(x+dx[i]>&&x+dx[i]<=m&&y+dy[i]>&&y+dy[i]<=n)
     if(a[x+dx[i]][y+dy[i]]) if(!vis[x+dx[i]][y+dy[i]]) dfs(x+dx[i],y+dy[i]);
     if(col[x][y]==) if(cr(x,y)) {     ans<<"D "<<x<<' '<<y<<"\n";
      ans<<"R "<<x<<' '<<y<<"\n";col[x][y]=;sum+=;}
 }
 int main()
 {
 cin>>m>>n;
 FOR(i,,m)
 FOR(j,,n)
 {
     cin>>x;
     a[i][j]=(x=='.');
 }
 FOR(i,,m)
 FOR(j,,n)
 if(a[i][j]&&!vis[i][j]) dfs(i,j);
 cout<<sum<<endl;
 cout<<ans.str();
 return ;
 }

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][]);