[高斯消元] POJ 2345 Central heating

Central heating

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 614		Accepted: 286

Description

Winter has come, but at the Ural State University heating is not turned on yet. There's one little problem: the University is heated only if all of the valves are opened. There are some technicians at the University. Each of them is responsible for one or more valves. There may be several technicians responsible for the same valve. When a technician gets an instruction to turn on the heating he goes round all of his valves and turns them. It means that if a valve was opened then he closes it, and if it was closed then he opens it. It is well known that every technician earns his money not in vain so it's impossible to replace any technician by any combination of other technicians. 
Your task is to determine who of the technicians is to get an instruction "to turn on the heating" in order to heat all the Ural State University. Note that there are N technicians and N valves at the University (1 <= N <= 250). 

Input

The first line of an input contains the number N. The next N lines contain lists of the valves in charge of each of the technicians. It means that a line number i + 1 contains numbers of the valves that the i-th technician is responsible for. Each list of valves is followed by –1.

Output

An output should contain a list of technicians' numbers sorted in ascending order. If several lists are possible, you should send to an output the shortest one. If it's impossible to turn on the heating at the University, you should write "No solution" .

Sample Input

4
1 2 -1
2 3 4 -1
2 -1
4 -1

Sample Output

1 2 3

Source

Ural State University Internal Contest October'2000 Students Session

原题大意： 有n个人，n个阀门。给n组数，每i组一-1结束，代表第i个人管理这组数中正数编号的阀门。

              问：能否确定几个人，使得所有阀门都开着。

解题思路： 不知道为啥是高斯消元，明明线性代数用的比较多。

              由于同一个人，开偶数次与不开是一样的，开奇数次与开1次是一样的。

              于是对于一个人，只有两种情况，不开和开一次。

              既然如此，我们可以把每个人的状态做成n维列向量，再做成增广矩阵，如题中所示。

             1  0  0  0  |  1

             1  1  1  0  |  1

             0  1  0  0  |  1

             0  1  0  1  |  1

             这样，解这个增广矩阵就可以了。

             值得一提的是，原文中有 it's impossible to replace any technician by any combination of other technicians. 

             什么意思呢，也就是说，这个系数矩阵分成向量后是线性无关的。

             也就是说，系数矩阵是n!必定有唯一的解。于是就可以无视原题中的无解和多解情况了。

#include<stdio.h>
#include<string.h>
int a[270][270],ans[270],n;
void swap(int *a,int *b)
  {
  	 int cnt=0,c[270],i;
  	 for(cnt=1;cnt<=n+1;++cnt)
	   {
	     c[cnt]=*a;
	     *a++=*(b+cnt-1);
       }
  	 for(i=1;i<=n+1;++i) *b++=c[i]; 
  }
void XOR(int col,int *a,int *b)
  {
  	 int i;
  	 for(i=col;i<=n+1;++i) *b++^=*a++;
  }
void init()
  {
  	 int col=0,x,i;
  	 for(i=1;i<=n;++i) a[i][n+1]=1;
  	 for(col=1;col<=n;++col) while(~scanf("%d",&x)&&x!=-1) a[x][col]=1;
  }
void solved()
  {
  	 int col,row,node;
  	 for(col=1;col<=n;++col)
  	   {
  	   	  node=0;
  	   	  for(row=col;row<=n;++row)
  	   	    {
  	   	       if(a[row][col]) 
  	   	         {
  	   	         	node=row;
  	   	         	break;
				 }
			}
		  if(node) swap(a[node]+1,a[col]+1);
		  for(row=col+1;row<=n;++row)
		    if(a[row][col]) XOR(col,a[col]+col,a[row]+col);
	   }
  }
void target()
  {
  	 int i,j;
  	 for(i=n;i>=1;--i)
  	   {
  	   	  ans[i]=a[i][n+1];
  	   	  for(j=n-1;j>=1;--j) a[j][n+1]^=(ans[i] & a[j][i]);
	   }
     int first=1;
     for(i=1;i<=n;++i) if(ans[i]) printf("%d ",i);
	 printf("
");
  }
int main()
  {
  	 int t,x,col=0;
  	 scanf("%d",&n);
  	 init();
  	 solved();
  	 target();
  	 return 0;
  }