POJ3185——The Water Bowls

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Description

The cows have a line of 20 water bowls from which they drink. The bowls can be either right-side-up (properly oriented to serve refreshing cool water) or upside-down (a position which holds no water). They want all 20 water bowls to be right-side-up and thus use their wide snouts to flip bowls.

Their snouts, though, are so wide that they flip not only one bowl but also the bowls on either side of that bowl (a total of three or -- in the case of either end bowl -- two bowls).

Given the initial state of the bowls (1=undrinkable, 0=drinkable -- it even looks like a bowl), what is the minimum number of bowl flips necessary to turn all the bowls right-side-up?

Input

Line 1: A single line with 20 space-separated integers

Output

Line 1: The minimum number of bowl flips necessary to flip all the bowls right-side-up (i.e., to 0). For the inputs given, it will always be possible to find some combination of flips that will manipulate the bowls to 20 0's.

Sample Input

0 0 1 1 1 0 0 1 1 0 1 1 0 0 0 0 0 0 0 0

Sample Output

Hint

Explanation of the sample:

Flip bowls 4, 9, and 11 to make them all drinkable:
0 0 1 1 1 0 0 1 1 0 1 1 0 0 0 0 0 0 0 0 [initial state]
0 0 0 0 0 0 0 1 1 0 1 1 0 0 0 0 0 0 0 0 [after flipping bowl 4]
0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 [after flipping bowl 9]
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 [after flipping bowl 11]

Source

USACO 2006 January Bronze

典型的高消解异或方程组，需要枚举自由变元。

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<iostream>
#include<algorithm>

using namespace std;

const int maxn=30;
int x[maxn];//存最后的解
int mat[maxn][maxn];
int free_x[maxn];//存自由变元

int Gauss(int equ,int var)
{
	int i,j,k;
	int free_index,free_num;
	int max_r,col,num;
	col=0;
	num=0;
	memset(x,0,sizeof(x));
	for(int i=0;i<var+1;i++)
    {
        x[i]=0;
        free_x[i]=0;
    }
	for(k=0;k<equ && col<var;k++,col++)
	{
		max_r=k;//找到col列上绝对值最大的那行(k行的下面那些行)
		for(i=k+1;i<equ;i++)
		{
			if(abs(mat[i][col]) > abs(mat[max_r][col]))
			  max_r=i;
		}
		if(max_r!=k)//找到后和第k行交换
		{
			for(j=k;j<var+1;j++)
			  swap(mat[k][j],mat[max_r][j]);
		}
		if(!mat[k][col])
        {
        	k--;
        	free_x[num++]=col;//如果第k行第col这个数为0，也就是系数为0，那他就是个自由变量
        	continue;
        }
        for(i=k+1;i<equ;i++)
        {
        	if(mat[i][col])
        	{
	        	for(j=col;j<var+1;j++)
	        	  mat[i][j]^=mat[k][j];//异或以后消去k+1行到最后一行的第col列的系数，形成上三角矩阵
	        }
        }
	}
	for(i=k;i<equ;i++)
	  if(mat[i][col])
	    return -1;//无解
	//接下来得枚举自由变元
	int all=1<<(var-k);//var-k个自由变元，那么有2的指数次个状态
	int res=0x3f3f3f3f;
	for(i=0;i<all;i++)
	{
		int cnt=0;
		int index=i;
		for(j=0;j<var-k;j++)
		{
			x[free_x[j]]=(index&1);
			if(index&1)//如果是1，则要反转，也就是要操作一次
			  cnt++;
			index>>=1;
		}
		//同时还要判断确定的变量
		for(j=k-1;j>=0;j--)
	    {
	     int temp=mat[j][var];
	     for(int l=j+1;l<var;l++)
	        if(mat[j][l])
              temp^=mat[j][l]*x[l];
         x[j]=temp;
         if(x[j])
          cnt++;
	     }
		if(cnt < res)
		  res = cnt;
	}
	return res;
}

int num[22];

int main()
{
	while(~scanf("%d",&num[0]))
	{
		memset(mat,0,sizeof(mat));
		for(int i=1;i<20;i++)
		  scanf("%d",&num[i]);
        for(int i=0;i<20;i++)
        {
       	  mat[i][i]=1;
       	  if(i>0)
           mat[i][i-1]=1;
      	  if(i<19)
   	       mat[i][i+1]=1;
        }
        for(int i=0;i<20;i++)
            mat[i][20]=num[i];
        int res=Gauss(20,20);
        printf("%d\n",res);
	}
	return 0;
}