zjuoj 3780 Paint the Grid Again

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3780

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). Then N 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 solution X is lexicographically smaller than Y if there exists an integer k, the first k - 1 operations of X and Y are the same. The k-th operation of X 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

2
2
XX
OX
2
XO
OX

Sample Output

R2 C1 R1
No solution

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Author: YU, Xiaoyao
Source: The 11th Zhejiang Provincial Collegiate Programming Contest

 

 

 

分析；

給定n*n的矩陣

有2個操做：

一、把一行變成X

二、把一列變成O

限制：每行（每列）只能變一次

給定結果圖，開始時圖無O,X，問最小操做步數（且字典序最小）

思路：

對於(i,j)這個格子，若如今塗的是 O，則去掉O這排，（讓這排都變成X便可）能夠直接認爲(i,j)是X

因此當某排的X攢滿n個時，就能夠去掉這排X

直接模擬便可

先把全部 全爲O或全爲X的 行和列預處理出來，放到一個棧裏

由於字典序最小，因此先處理列再處理行，第i列 用i+n表示， 第i行用i表示
而後給棧排個序，這樣就獲得處理當前狀況的順序， 入個隊列，而後一個個去掉就能夠了。

 

 

AC代碼：

  1 #include<stdio.h>
  2 #include<iostream>
  3 #include<algorithm>
  4 #include<string.h>
  5 #include<math.h>
  6 #include<vector>
  7 #include<queue>
  8 #include<set>
  9 using namespace std;
 10 #define N 1005
 11 vector<int>ans;
 12 char mp[N][N];
 13 int n, h[N], l[N];
 14 int yes[N];
 15 int Stack[N], Top;
 16 
 17 void init(){
 18     ans.clear();
 19     memset(yes, 0, sizeof yes);
 20     memset(h, 0, sizeof h);
 21     memset(l, 0, sizeof l);
 22     Top = 0;
 23 }
 24 bool cmp(int a,int b){return a>b;}
 25 //0-n-1 表示列 n-2n-1 表示行
 26 void work(){
 27     sort(Stack, Stack+Top, cmp);
 28     queue<int>q;
 29     int i, j;
 30     for(int i = 0; i < Top; i++){
 31         q.push(Stack[i]), ans.push_back(Stack[i]); yes[Stack[i]]=-1;
 32     }
 33     Top = 0;
 34     while(!q.empty()){
 35         int u = q.front(); q.pop();
 36         Top = 0;
 37         if(u<n)
 38             for(j = 0; j < n; j++)
 39             {
 40                 mp[j][u] = 'X';
 41                 h[j]++;
 42                 if(yes[j+n]!=-1 && h[j]==n)Stack[Top++] = j+n;
 43             }
 44         else {
 45             u-=n;
 46             for(j = 0; j < n; j++)
 47             {
 48                 mp[u][j] = 'O';
 49                 l[j]++;
 50                 if(yes[j]!=-1 && l[j]==n)Stack[Top++] = j;
 51             }
 52         }
 53             sort(Stack, Stack+Top, cmp);
 54             for(i = 0; i < Top; i++)q.push(Stack[i]), yes[Stack[i]] = -1, ans.push_back(Stack[i]);
 55     }
 56     for(int i = 0; i < 2*n; i++)if(yes[i]==0){puts("No solution");return;}
 57     for(int i = ans.size()-1; i>=0; i--){
 58         int u = ans[i];
 59         if(u>=n)printf("R"), u-=n;
 60         else printf("C");
 61         printf("%d",u+1);
 62         i ? printf(" ") : puts("");
 63     }
 64 }
 65 int main(){
 66     int T;scanf("%d",&T);
 67     int i, j;
 68     while(T--){
 69         scanf("%d",&n);
 70         init();
 71         for(i=0;i<n;i++)scanf("%s",mp[i]);
 72         for(i=0;i<n;i++)
 73         {
 74             for(j = 0; j<n; j++)if(mp[i][j]=='X')h[i]++;
 75             if(h[i]==n) Stack[Top++] = i+n;
 76             else if(h[i]==0) yes[i+n] = -1;
 77         }
 78         for(i=0;i<n;i++)
 79         {
 80             for(j = 0; j<n; j++)if(mp[j][i]=='O')l[i]++;
 81             if(l[i]==n) Stack[Top++] = i;
 82             else if(l[i]==0) yes[i] = -1;
 83         }
 84         if(Top==0){puts("No solution");continue;}
 85         work();
 86     }
 87     return 0;
 88 }
 89 /*
 90 99
 91 1
 92 O
 93 3
 94 OOO
 95 OOO
 96 OOO
 97 
 98 2
 99 XX
100 OX
101 2
102 XO
103 OX
104 
105 
106 */

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