hdu 5755 2016 Multi-University Training Contest 3 Gambler Bo 高斯消元模3同余方程

http://acm.hdu.edu.cn/showproblem.php?pid=5755

题意：一个N*M的矩阵，改变一个格子，本身+2，四周+1.同时mod 3;问操作多少次，矩阵变为全0.输出次数和具体位置

由于影响是相互的，所以增广矩阵的系数a[t][t+1] 或者是 a[t+1][t]均可；只需注意往结果中添加位置时，x[i]表示要操作x[i]次，所以位置要重复添加x[i]次；

高斯消元时间复杂度为O（N³M³）;

 #pragma comment(linker, "/STACK:1024000000,1024000000")
 #include<bits/stdc++.h>
 using namespace std;
 #define rep0(i,l,r) for(int i = (l);i < (r);i++)
 #define rep1(i,l,r) for(int i = (l);i <= (r);i++)
 #define rep_0(i,r,l) for(int i = (r);i > (l);i--)
 #define rep_1(i,r,l) for(int i = (r);i >= (l);i--)
 #define MS0(a) memset(a,0,sizeof(a))
 #define MS1(a) memset(a,-1,sizeof(a))
 #define MSi(a) memset(a,0x3f,sizeof(a))
 #define pb push_back
 #define A first
 #define B second
 #define MK make_pair
 #define inf 0x3f3f3f3f
 #define eps 1e-8
 #define zero(x) (((x)>0?(x):-(x))<eps)
 #define bitnum(a) __builtin_popcount(a)
 #define lowbit(x) (x&(-x))
 #define clear0 (0xFFFFFFFE)
 #define mod 3
 #define K(x) ((x)*(x))
 typedef pair<int,int> PII;
 typedef long long ll;
 typedef unsigned long long ull;
 template<typename T>
 void read1(T &m)
 {
     T x = ,f = ;char ch = getchar();
     while(ch <'' || ch >''){ if(ch == '-') f = -;ch=getchar(); }
     while(ch >= '' && ch <= ''){ x = x* + ch - '';ch = getchar(); }
     m = x*f;
 }
 template<typename T>
 void read2(T &a,T &b){read1(a);read1(b);}
 template<typename T>
 void read3(T &a,T &b,T &c){read1(a);read1(b);read1(c);}
 template<typename T>
 void out(T a)
 {
     if(a>) out(a/);
     putchar(a%+'');
 }
 inline ll gcd(ll a,ll b){ return b == ? a: gcd(b,a%b); }
 inline ll lcm(ll a,ll b){ return a/gcd(a,b)*b; }
 const int maxn = ;
 int a[maxn][maxn];
 int x[maxn<<];

 void init(int n,int m)
 {
     rep0(i,,n) rep0(j,,m){
         int t = i*m + j;
         a[t][t] = ;
         if(i > ) a[t - m][t] = ;
         if(i < n-) a[t + m][t] = ;
         if(j > ) a[t-][t] = ;
         if(j < m-) a[t+][t] = ;
     }
 }

 int equ, var;
 int Gauss()
 {
     int mx, row, col;
     for(row = , col = ; row < equ && col < var; row++, col++){
         mx = row;
         for(int i = row+;i < equ;i++){
             if(abs(a[i][col]) > abs(a[mx][col])) mx = i;
         }
         if(a[mx][col] == ) {
             row--;
             continue;
         }
         if(mx != row){
             for(int j = col;j < var + ;j++)
                 swap(a[row][j], a[mx][j]);
         }
         for(int i = row+;i < equ;i++){
             if(a[i][col]){
                 int LCM = lcm(abs(a[i][col]), abs(a[row][col]));
                 int ta = LCM/abs(a[i][col]), tb = LCM/abs(a[row][col]);
                 if(a[i][col] * a[row][col] < ) tb = -tb;
                 for(int j = col; j < var+;j++)
                     a[i][j] = ((a[i][j]*ta - a[row][j]*tb)%mod+mod)%mod;
             }
         }
     }

     for(int i = var-; i >= ; i--){
         if(a[i][i] == ) continue;
         int tmp = a[i][var];
         for(int j = i+;j < var;j++){
             if(a[i][j]){
                 tmp -= a[i][j]*x[j];
                 tmp = (tmp%mod + mod)% mod;
             }
         }
         x[i] = (tmp*a[i][i])% mod;
     }
     return ;
 }
 vector<int> vec;
 int main()
 {
     //freopen("data.txt","r",stdin);
     //freopen("out.txt","w",stdout);
     int T, kase = ;
     scanf("%d",&T);
     while(T--){
         int n, m, t;
         read2(n,m);
         equ = n*m, var = n*m;
         MS0(a);
         rep0(i,,n) rep0(j,,m) read1(t), a[i*m+j][var] = (mod-t)%mod;
         init(n,m);

         Gauss();
         vec.clear();
         rep0(i,,var)  while(x[i]--) vec.pb(i);
         out(vec.size());puts("");
         rep0(i,,vec.size()) printf("%d %d\n",vec[i]/m + , vec[i]%m + );
     }
     return ;
 }