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 }
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 }
输出黑科技sstream orz
#include<sstream> x; x<<.....; cout<<x.str();
输入黑科技
scanf("%s",&string[i][1]);
