zoj 3780 Paint the Grid Again (拓扑排序)

Paint the Grid Again

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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Leo has a grid with N × N cells. 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 integer T indicating the number of test cases. For each test case:

The first line contains an integer N (1 <= N <= 500). ThenN lines follow. Each line contains a string with N characters. Each character is either 'X' (black) or 'O' (white) indicates the color of the cells should 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 solutionX is lexicographically smaller than Y if there exists an integerk, the first k - 1 operations of X and Y are the same. The k-th operation ofX is smaller than the k-th in Y. 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 R1

No solution

题意：

将一面墙刷成给出的图形，你可以将一排刷成X或者将一列刷成O，问字典序最小的方案是什么。

思路：

拓扑排序的简单应用，好久没见拓扑排序的题了，竟然都没看出来。衰~

将每行每列看做点，根据每个点为X或者O来建图，然后dfs或者用优先队列就能找到字典序最小的方案了。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <vector>#include <queue>#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<int>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<int, vector<int>, greater<int> > 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<v[x].size();j++)            {                num[v[x][j]]--;            }        }        if(i<tot) printf("No solution\n");        else        {            for(i=1;i<=tot;i++)            {                if(i>1) printf(" ");                if(res[i]<=n) printf("C%d",res[i]);                else printf("R%d",res[i]-n);            }            printf("\n");        }    }    return 0;}