Mathematical---Gaussian elimination POJ 1830

Switching problems problem ' s link:http://poj.org/problem?id=1830

Mean:

Slightly

Analyse:

Slightly augmented matrix: Con[i][j]: If the Operation J,i state change is con[i][j]=1, otherwise con[i][j]=0.

The last augmented matrix should be N (n+1), the last column: The whole state of Bikai, if the same is 0, if the difference is 1;

There are three kinds of final solutions:
1. No solution, there is a row in front of the number of n is 0, the value of the n+1 is not 0;
2. There is no 1th case, the existence of X-line value is all 0, then the number of solutions is 2^x;
3, No 1, 22 cases appear, the only solution, Output 1.

Time Complexity:o (n)

Source Code:

/** This code was made by crazyacking* verdict:accepted* submission date:2015-06-17-22.36* time:0ms* memory:137kb*/#include<queue>#include<cstdio>#include<Set>#include<string>#include<stack>#include<cmath>#include<climits>#include<map>#include<cstdlib>#include<iostream>#include<vector>#include<algorithm>#include<cstring>#defineLL Long Long#defineULL unsigned long Longusing namespacestd;Const intp= -;intCon[p][p];intN;intBeg[p],fin[p];intfunction () {intI,j,k,t,row,col,temp,count=0;  for(row=col=1; row<=n&&col<=n; row++,col++)      {            if(con[row][col]==0)            {                   for(i=row+1; i<=n; i++)                  {                        if(con[i][col]!=0)                        {                               for(j=1; j<=n+1; J + +) {swap (con[row][j],con[i][j]); }                               Break; }                  }            }            if(con[row][col]==0) {row--; Continue; }             for(k=1; k<=n; k++)             {                  if(con[k][col]!=0&&k!=row) {Temp=-(con[k][col]/Con[row][col]);  for(T=col; t<=n+1; t++) {Con[k][t]= (Temp*con[row][t]) +Con[k][t]; }                  }            }      }       for(K=row; k<n+1; k++)            if(con[k][n+1]!=0) {return 0; } if(row==n+1) {return 1; } Else      {            return 1<< (n-row+1); }}intMain () {intT,i,j,x,y; scanf ("%d",&T);  while(t--) {scanf ("%d",&N);  for(i=1; i<=n; i++) {scanf ("%d",&Beg[i]); }             for(i=1; i<=n; i++) {scanf ("%d",&Fin[i]); } scanf ("%d%d",&x,&y); Memset (Con,0,sizeof(Con));  while(x!=0&&y!=0) {Con[y][x]=1; scanf ("%d%d",&x,&y); }             for(i=1; i<=n; i++) {Con[i][i]=1; }             for(i=1; i<=n; i++)            {                  if(beg[i]==Fin[i]) {Con[i][n+1]=0; }                  Else{con[i][n+1]=1; }            }            intPP =function (); if(PP) {printf ("%d\n", pp); }            Else{printf ("Oh,it ' s impossible~!! \ n"); }      }}

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Mathematical---Gaussian elimination POJ 1830