Gambler Bo (Gauss elimination element for special solution)

For each point in the diagram, assume that you click XI * m + j and then each point has so that for each point you can enumerate an equation, n*m a point solution n*m an unknown. It can be solved by using Gauss elimination element.

The problem is that there may be more than one special, so we need to solve the kindergarten solution. And then that one solved I can't read.

#include <bits/stdc++.h>using namespacestd;Const intMAXN = ** *;intN, M, CNT;intid[ *][ *], data[ *][ *], A[MAXN][MAXN], X[MAXN];intgcdintAintb) {    returnB?GCD (b, a%b): A;}intLcmintAintb) {    returnA/GCD (A, b) *b;}voidinit () {memset (x,0,sizeof(x)); Memset (A,0,sizeof(a));  for(inti =1; I <= N; i + +){         for(intj =1; J <= M; J + +) {a[id[i][j]][cnt]= (3-DATA[I][J])%3; A[ID[I][J]][ID[I][J]]=2; if(I >1) A[id[i][j]][id[i-1][J]] =1; if(J >1) A[id[i][j]][id[i][j-1]] =1; if(I < n) a[id[i][j]][id[i +1][J]] =1; if(J < m) A[id[i][j]][id[i][j +1]] =1; }    }}voidGaussi () { for(inti =1; I < CNT; i + +){        inttop =i;  for(intj = i +1; J < CNT; J + +) Top= ABS (A[j][i]) > abs (A[top][i])?J:top; if(A[top][i]) { for(intj = i; J <= CNT; J + +) Swap (A[top][j], a[i][j]);  for(intj = i +1; J < CNT; J + +)            if(A[j][i]) {intD =LCM (A[j][i], a[i][i]); intX1 = D/a[j][i], x2 = d/A[i][i];  for(intK = i; K <= CNT; K + +) A[j][k]= ((a[j][k] * x1-a[i][k] * x2)%3+3) %3; }        }    }    intAns =0;  for(inti = cnt-1; i >0; I--) {X[i]=a[i][cnt];  for(intj = i +1; J < CNT; J + +) X[i]= ((X[i]-a[i][j] * x[j])%3+3)%3; X[i]= a[i][i] * X[i]%3; Ans+=X[i]; } printf ("%d\n", ans);  for(inti =1; I < CNT; i + +){         while(X[i]) {printf ("%d%d\n", (I-1)/M +1, (I-1)%m +1); X[i]--; }    }}intMain () {intT; scanf ("%d",&T);  while(T--) {CNT=1; scanf ("%d%d",&n,&m);  for(inti =1; I <= N; i + +)             for(intj =1; J <= M; J + +) scanf ("%d", &data[i][j]), id[i][j] = cnt++;        Init ();    Gaussi (); }    return 0 ;}

Gambler Bo (Gauss elimination element for special solution)