Zoj 3780 Paint the Grid Again (topological sorting)

Paint the Grid Again Time Limit: 2 Seconds Memory Limit: 65536 KB

Leo has a gridN×NCells. He wants to paint each cell with a specific color (either black or white ).

Leo has a magical brush which can paint any row with black color, or any column with white color. each time he uses the brush, the previous color of cells will be covered by the new color. since the magic of the brush is limited, each row and each column can only be painted at most once. the cells were painted in some other color (neither black nor white) initially.

Please write a program to find out the way to paint the grid.

Input

There are multiple test cases. The first line of input contains an integerTIndicating the number of test cases. For each test case:

The first line contains an integerN(1 <=N<= 500). ThenNLines follow. Each line contains a stringNCharacters. Each character is either 'X' (black) or 'O' (white) indicates the color of the cells shoshould be painted to, after Leo finished his painting.

Output

For each test case, output "No solution" if it is impossible to find a way to paint the grid.

Otherwise, output the solution with minimum number of painting operations. each operation is either "R #" (paint in a row) or "C #" (paint in a column), "#" is the index (1-based) of the row/column. use exactly one space to separate each operation.

Among all possible solutions, you should choose the lexicographically smallest one. A solutionXIs lexicographically smallerYIf there exists an integerK, The firstK-1 operationsXAndYAre the same. The k-th operationXIs smaller than the k-th inY. The operation in a column is always smaller than the operation in a row. If two operations have the same type, the one with smaller index of row/column is the lexicographically smaller one.

Sample Input

22XXOX2XOOX

Sample Output
R2 C1 R1No solution

Q: You can use a wall to generate a line of X or a column of O to ask what the solution with the smallest Lexicographic Order is.
Idea: the simple application of topological sorting. I haven't seen the topic of topological sorting for a long time. Decline ~ Think of each column of each row as a vertex, and create a graph based on X or O for each vertex. Then, the dfs or priority queue can be used to find the solution with the smallest Lexicographic Order.
Code:

#include 
 
  #include 
  
   #include 
   
    #include 
    
     #include 
     
      #define maxn 1005using namespace std;int n,m,ans,tot,flag,cnt;bool vis[maxn];int app[maxn],num[maxn],res[maxn];char mp[maxn][maxn];vector
      
       v[maxn];void build(){ int i,j,t; tot=0; memset(app,0,sizeof(app)); memset(num,0,sizeof(num)); for(i=1; i<=n; i++) { flag=0; for(j=1; j<=n; j++) { if(mp[i][j]=='X') { flag=1; break ; } } if(flag) { tot++,app[n+i]=1; } } for(j=1; j<=n; j++) { flag=0; for(i=1; i<=n; i++) { if(mp[i][j]=='O') { flag=1; break ; } } if(flag) { tot++,app[j]=1; } } for(i=1; i<=n; i++) { for(j=1; j<=n; j++) { if(mp[i][j]=='X') { if(app[j]) { num[n+i]++; v[j].push_back(n+i); } } else { if(app[n+i]) { num[j]++; v[n+i].push_back(j); } } } }}int main(){ int i,j,t; scanf("%d",&t); while(t--) { scanf("%d",&n); for(i=1; i<=n; i++) { scanf("%s",mp[i]+1); v[i].clear();v[i+n].clear(); } build(); priority_queue
       
        , greater
        
          > q; memset(vis,0,sizeof(vis)); for(i=1;i<=tot;i++) { for(j=1;j<=n+n;j++) { if(app[j]&&num[j]==0&&!vis[j]) vis[j]=1,q.push(j); } if(q.empty()) break ; int x=q.top(); res[i]=x; q.pop(); for(j=0;j
         
          1) printf(" "); if(res[i]<=n) printf("C%d",res[i]); else printf("R%d",res[i]-n); } printf("\n"); } } return 0;}