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）;
1 #pragma comment(linker, "/STACK:1024000000,1024000000") 2 #include<bits/stdc++.h> 3 using namespace std; 4 #define rep0(i,l,r) for(int i = (l);i < (r);i++) 5 #define rep1(i,l,r) for(int i = (l);i <= (r);i++) 6 #define rep_0(i,r,l) for(int i = (r);i > (l);i--) 7 #define rep_1(i,r,l) for(int i = (r);i >= (l);i--) 8 #define MS0(a) memset(a,0,sizeof(a)) 9 #define MS1(a) memset(a,-1,sizeof(a)) 10 #define MSi(a) memset(a,0x3f,sizeof(a)) 11 #define pb push_back 12 #define A first 13 #define B second 14 #define MK make_pair 15 #define inf 0x3f3f3f3f 16 #define eps 1e-8 17 #define zero(x) (((x)>0?(x):-(x))<eps) 18 #define bitnum(a) __builtin_popcount(a) 19 #define lowbit(x) (x&(-x)) 20 #define clear0 (0xFFFFFFFE) 21 #define mod 3 22 #define K(x) ((x)*(x)) 23 typedef pair<int,int> PII; 24 typedef long long ll; 25 typedef unsigned long long ull; 26 template<typename T> 27 void read1(T &m) 28 { 29 T x = 0,f = 1;char ch = getchar(); 30 while(ch <'0' || ch >'9'){ if(ch == '-') f = -1;ch=getchar(); } 31 while(ch >= '0' && ch <= '9'){ x = x*10 + ch - '0';ch = getchar(); } 32 m = x*f; 33 } 34 template<typename T> 35 void read2(T &a,T &b){read1(a);read1(b);} 36 template<typename T> 37 void read3(T &a,T &b,T &c){read1(a);read1(b);read1(c);} 38 template<typename T> 39 void out(T a) 40 { 41 if(a>9) out(a/10); 42 putchar(a%10+'0'); 43 } 44 inline ll gcd(ll a,ll b){ return b == 0? a: gcd(b,a%b); } 45 inline ll lcm(ll a,ll b){ return a/gcd(a,b)*b; } 46 const int maxn = 907; 47 int a[maxn][maxn]; 48 int x[maxn<<2]; 49 50 void init(int n,int m) 51 { 52 rep0(i,0,n) rep0(j,0,m){ 53 int t = i*m + j; 54 a[t][t] = 2; 55 if(i > 0) a[t - m][t] = 1; 56 if(i < n-1) a[t + m][t] = 1; 57 if(j > 0) a[t-1][t] = 1; 58 if(j < m-1) a[t+1][t] = 1; 59 } 60 } 61 62 int equ, var; 63 int Gauss() 64 { 65 int mx, row, col; 66 for(row = 0, col = 0; row < equ && col < var; row++, col++){ 67 mx = row; 68 for(int i = row+1;i < equ;i++){ 69 if(abs(a[i][col]) > abs(a[mx][col])) mx = i; 70 } 71 if(a[mx][col] == 0) { 72 row--; 73 continue; 74 } 75 if(mx != row){ 76 for(int j = col;j < var + 1;j++) 77 swap(a[row][j], a[mx][j]); 78 } 79 for(int i = row+1;i < equ;i++){ 80 if(a[i][col]){ 81 int LCM = lcm(abs(a[i][col]), abs(a[row][col])); 82 int ta = LCM/abs(a[i][col]), tb = LCM/abs(a[row][col]); 83 if(a[i][col] * a[row][col] < 0) tb = -tb; 84 for(int j = col; j < var+1;j++) 85 a[i][j] = ((a[i][j]*ta - a[row][j]*tb)%mod+mod)%mod; 86 } 87 } 88 } 89 90 for(int i = var-1; i >= 0; i--){ 91 if(a[i][i] == 0) continue; 92 int tmp = a[i][var]; 93 for(int j = i+1;j < var;j++){ 94 if(a[i][j]){ 95 tmp -= a[i][j]*x[j]; 96 tmp = (tmp%mod + mod)% mod; 97 } 98 } 99 x[i] = (tmp*a[i][i])% mod; 100 } 101 return 0; 102 } 103 vector<int> vec; 104 int main() 105 { 106 //freopen("data.txt","r",stdin); 107 //freopen("out.txt","w",stdout); 108 int T, kase = 1; 109 scanf("%d",&T); 110 while(T--){ 111 int n, m, t; 112 read2(n,m); 113 equ = n*m, var = n*m; 114 MS0(a); 115 rep0(i,0,n) rep0(j,0,m) read1(t), a[i*m+j][var] = (mod-t)%mod; 116 init(n,m); 117 118 Gauss(); 119 vec.clear(); 120 rep0(i,0,var) while(x[i]--) vec.pb(i); 121 out(vec.size());puts(""); 122 rep0(i,0,vec.size()) printf("%d %d ",vec[i]/m + 1, vec[i]%m + 1); 123 } 124 return 0; 125 }