Codeforces Round #684 (Div. 2) C1. Binary Table (Easy Version) 模拟
This is the easy version of the problem. The difference between the versions is in the number of possible operations that can be made. You can make hacks if and only if you solved both versions of the problem.
You are given a binary table of size n×m. This table consists of symbols 0 and 1.
You can make such operation: select 3 different cells that belong to one 2×2 square and change the symbols in these cells (change 0 to 1 and 1 to 0).
Your task is to make all symbols in the table equal to 0. You are allowed to make at most 3nm operations. You don’t need to minimize the number of operations.
It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤t≤5000) — the number of test cases. The next lines contain descriptions of test cases.
The first line of the description of each test case contains two integers n, m (2≤n,m≤100).
Each of the next n lines contains a binary string of length m, describing the symbols of the next row of the table.
It is guaranteed that the sum of nm for all test cases does not exceed 20000.
Output
For each test case print the integer k (0≤k≤3nm) — the number of operations.
In the each of the next k lines print 6 integers x1,y1,x2,y2,x3,y3 (1≤x1,x2,x3≤n,1≤y1,y2,y3≤m) describing the next operation. This operation will be made with three cells (x1,y1), (x2,y2), (x3,y3). These three cells should be different. These three cells should belong into some 2×2 square.
Example
inputCopy
5
2 2
10
11
3 3
011
101
110
4 4
1111
0110
0110
1111
5 5
01011
11001
00010
11011
10000
2 3
011
101
outputCopy
1
1 1 2 1 2 2
2
2 1 3 1 3 2
1 2 1 3 2 3
4
1 1 1 2 2 2
1 3 1 4 2 3
3 2 4 1 4 2
3 3 4 3 4 4
4
1 2 2 1 2 2
1 4 1 5 2 5
4 1 4 2 5 1
4 4 4 5 3 4
2
1 3 2 2 2 3
1 2 2 1 2 2
Note
In the first test case, it is possible to make only one operation with cells (1,1), (2,1), (2,2). After that, all symbols will be equal to 0.
In the second test case:
operation with cells (2,1), (3,1), (3,2). After it the table will be:
011
001
000
operation with cells (1,2), (1,3), (2,3). After it the table will be:
000
000
000
In the fifth test case:
operation with cells (1,3), (2,2), (2,3). After it the table will be:
010
110
operation with cells (1,2), (2,1), (2,2). After it the table will be:
000
000
同学写的带模拟。。。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <queue>
#include <functional>
#include <vector>
#include <stack>
#include <set>
#define int long long
using namespace std;
typedef long long ll;
const int maxn=2e5+50;
const int inf=0x3f3f3f3f;
const int MOD=1e9+7;
const int HASH=131;
int mp[200][200];
vector<int> ans[maxn];
signed main()
{
int t;
cin>>t;
while(t--)
{
for(int i=1;i<maxn;i++)
ans[i].clear();
int cnt=0;
int n,m;
cin>>n>>m;
string s[107];
for(int i=1;i<=n;i++)
cin>>s[i];
for(int i=1;i<=n;i++)
{
for(int j=0;j<m;j++)
{
mp[i][j+1]=s[i][j]-'0';
}
}
for(int i=1;i<=n;i++)
{
if(i!=n)
{
for(int j=1;j<=m;j++)
{
if(j!=m)
{
if(mp[i][j]!=0)
{
mp[i][j]^=1;
mp[i+1][j]^=1;
mp[i+1][j+1]^=1;
ans[++cnt].push_back(i);
ans[cnt].push_back(j);
ans[cnt].push_back(i+1);
ans[cnt].push_back(j);
ans[cnt].push_back(i+1);
ans[cnt].push_back(j+1);
}
}
else
{
if(mp[i][j]!=0)
{
mp[i][j]^=1;
mp[i+1][j]^=1;
mp[i+1][j-1]^=1;
ans[++cnt].push_back(i);
ans[cnt].push_back(j);
ans[cnt].push_back(i+1);
ans[cnt].push_back(j);
ans[cnt].push_back(i+1);
ans[cnt].push_back(j-1);
}
}
}
}
else
{
for(int j=1;j<=m;j++)
{
if(j!=m)
{
if(mp[i][j]!=0)
{
mp[i][j]^=1;
ans[++cnt].push_back(i);
ans[cnt].push_back(j);
ans[cnt].push_back(i-1);
ans[cnt].push_back(j);
ans[cnt].push_back(i-1);
ans[cnt].push_back(j+1);
mp[i][j+1]^=1;
ans[++cnt].push_back(i);
ans[cnt].push_back(j+1);
ans[cnt].push_back(i-1);
ans[cnt].push_back(j);
ans[cnt].push_back(i-1);
ans[cnt].push_back(j+1);
}
}
else
{
if(mp[i][j]!=0)
{
ans[++cnt].push_back(i);
ans[cnt].push_back(j);
ans[cnt].push_back(i);
ans[cnt].push_back(j-1);
ans[cnt].push_back(i-1);
ans[cnt].push_back(j);
ans[++cnt].push_back(i);
ans[cnt].push_back(j);
ans[cnt].push_back(i-1);
ans[cnt].push_back(j-1);
ans[cnt].push_back(i-1);
ans[cnt].push_back(j);
ans[++cnt].push_back(i);
ans[cnt].push_back(j);
ans[cnt].push_back(i-1);
ans[cnt].push_back(j-1);
ans[cnt].push_back(i);
ans[cnt].push_back(j-1);
mp[i][j]^=1;
}
}
}
}
}
cout<<cnt<<endl;
for(int i=1;i<=cnt;i++)
{
cout<<ans[i][0]<<' '<<ans[i][1]<<' '<<ans[i][2]<<' '<<ans[i][3]<<' '<<ans[i][4]<<' '<<ans[i][5]<<endl;
}
}
}