2016多校第三场 HDU 5755

一道明显的高斯消元模板题。。赛上sb的觉得 n3∗m3 会炸没敢写，推了半天最后一题的积分。。。。

建边也很简单，都是套路，模板。。

有的读者可能觉得程序写的有点问题，建边应该是别的点对当前点的贡献，但是因为这个题的性质，这个问题可以不考虑

//
//  Created by Running Photon
//  Copyright (c) 2015 Running Photon. All rights reserved.
//
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define ALL(x) x.begin(), x.end()
#define INS(x) inserter(x, x,begin())
#define ll long long
#define CLR(x) memset(x, 0, sizeof x)
using namespace std;
const int inf = 0x3f3f3f3f;
const int MOD = 3;
const int maxn = 1e6 + 10;
const int maxv = 1e3 + 10;
const double eps = 1e-9;

int a[maxv][maxv];
int x[maxv];
int mp[905];
int all;
void print() {
    for(int i = 0; i < all; i++) {
        for(int j = 0; j <= all; j++) {
            printf("%d ", a[i][j]);
        }
        puts("");
    }
    puts("");
}
int Gauss(int equ, int var) {
    int row, max_r, col;
    // print();
    for(row = 0, col = 0; col < var && row < equ; col++, row++) {
        max_r = row;
        for(int i = row + 1; i < equ; i++) {
            if(abs(a[i][col]) > abs(a[max_r][col])) {
                max_r = i;
            }
        }
        if(!a[max_r][col]) {
            row--;
            continue;
        }
        if(row != max_r) {
            for(int j = 0; j <= var; j++) {
                swap(a[row][j], a[max_r][j]);
            }
        }
        for(int i = row + 1; i < equ; i++) {
            if(a[i][col]) {
                int lcm = a[i][col] / __gcd(a[i][col], a[row][col]) * a[row][col];
                int t1 = lcm / a[i][col];
                int t2 = lcm / a[row][col];
                for(int j = col; j <= var; j++) {
                    a[i][j] = ((a[i][j] * t1 - a[row][j] * t2) % MOD + MOD) % MOD;
                }
            }
        }
    }
    // print();
    // for(int i = row; i < equ; i++)
    //     if(a[i][var]) continue;
    for(int i = var - 1; i >= 0; i--) {
        x[i] = a[i][var];
        for(int j = i + 1; j < var; j++) {
            //
            x[i] = ((x[i] - a[i][j] * x[j]) % MOD + MOD) % MOD;
        }
        //mod 3下的逆元就是其本身
        x[i] = a[i][i] * x[i] % MOD;
        // printf("x[%d] = %d\n", i, x[i]);

    }
}
bool check(int u) {
    return u >= 0 && u < all;
}
int main() {
#ifdef LOCAL
    freopen("C:\\Users\\Administrator\\Desktop\\in.txt", "r", stdin);
    freopen("C:\\Users\\Administrator\\Desktop\\out.txt","w",stdout);
#endif
//    ios_base::sync_with_stdio(0);
    int T;
    scanf("%d", &T);
    while(T--) {
        int n, m;
        scanf("%d%d", &n, &m);
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                scanf("%d", &mp[i*m+j]);
            }
        }
        all = n * m;
        CLR(a);
        for(int i = 0; i < all; i++) {
            int u = i - m;
            int d = i + m;
            int l = i - 1;
            int r = i + 1;
            a[i][i] = 2;
            a[i][all] = (3 - mp[i]) % MOD;
            if(check(u)) a[i][u] = 1;
            if(check(d)) a[i][d] = 1;
            if(check(l) && i % m != 0) a[i][l] = 1;
            if(check(r) && i % m != m - 1) a[i][r] = 1;
        }
        Gauss(all, all);
        std::vector<int> ans;
        for(int i = 0; i < all; i++) {
            while(x[i]) {
                x[i]--;
                ans.push_back(i);
            }
        }
        cout << ans.size() << endl;
        for(int i = 0; i < ans.size(); i++) {
            int r = ans[i] / m;
            int c = ans[i] % m;
            printf("%d %d\n", r+1, c+1);
        }
    }

    return 0;
}