http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3780
Paint the Grid Again
Time Limit: 2 Seconds Memory Limit: 65536 KB
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
Author: YU, Xiaoyao
Source: The 11th Zhejiang Provincial Collegiate Programming Contest
分析；
给定n*n的矩阵
有2个操作：
1、把一行变成X
2、把一列变成O
限制：每行（每列）只能变一次
给定结果图，开始时图无O,X，问最小操作步数（且字典序最小）
思路：
对于(i,j)这个格子，若现在涂的是 O，则去掉O这排，（让这排都变成X即可）可以直接认为(i,j)是X
所以当某排的X攒满n个时，就可以去掉这排X
直接模拟即可
先把所有 全为O或全为X的 行和列预处理出来，放到一个栈里
因为字典序最小，所以先处理列再处理行，第i列 用i+n表示， 第i行用i表示
然后给栈排个序，这样就得到处理当前情况的顺序， 入个队列，然后一个个去掉就可以了。
AC代码：
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<math.h>
#include<vector>
#include<queue>
#include<set>
using namespace std;
#define N 1005
vector<int>ans;
char mp[N][N];
int n, h[N], l[N];
int yes[N];
int Stack[N], Top; void init(){
ans.clear();
memset(yes, , sizeof yes);
memset(h, , sizeof h);
memset(l, , sizeof l);
Top = ;
}
bool cmp(int a,int b){return a>b;}
//0-n-1 表示列 n-2n-1 表示行
void work(){
sort(Stack, Stack+Top, cmp);
queue<int>q;
int i, j;
for(int i = ; i < Top; i++){
q.push(Stack[i]), ans.push_back(Stack[i]); yes[Stack[i]]=-;
}
Top = ;
while(!q.empty()){
int u = q.front(); q.pop();
Top = ;
if(u<n)
for(j = ; j < n; j++)
{
mp[j][u] = 'X';
h[j]++;
if(yes[j+n]!=- && h[j]==n)Stack[Top++] = j+n;
}
else {
u-=n;
for(j = ; j < n; j++)
{
mp[u][j] = 'O';
l[j]++;
if(yes[j]!=- && l[j]==n)Stack[Top++] = j;
}
}
sort(Stack, Stack+Top, cmp);
for(i = ; i < Top; i++)q.push(Stack[i]), yes[Stack[i]] = -, ans.push_back(Stack[i]);
}
for(int i = ; i < *n; i++)if(yes[i]==){puts("No solution");return;}
for(int i = ans.size()-; i>=; i--){
int u = ans[i];
if(u>=n)printf("R"), u-=n;
else printf("C");
printf("%d",u+);
i ? printf(" ") : puts("");
}
}
int main(){
int T;scanf("%d",&T);
int i, j;
while(T--){
scanf("%d",&n);
init();
for(i=;i<n;i++)scanf("%s",mp[i]);
for(i=;i<n;i++)
{
for(j = ; j<n; j++)if(mp[i][j]=='X')h[i]++;
if(h[i]==n) Stack[Top++] = i+n;
else if(h[i]==) yes[i+n] = -;
}
for(i=;i<n;i++)
{
for(j = ; j<n; j++)if(mp[j][i]=='O')l[i]++;
if(l[i]==n) Stack[Top++] = i;
else if(l[i]==) yes[i] = -;
}
if(Top==){puts("No solution");continue;}
work();
}
return ;
}
/*
99
1
O
3
OOO
OOO
OOO 2
XX
OX
2
XO
OX */