poj3185 高斯消元

The Water Bowls

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5329		Accepted: 2081

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

3

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]

题意：

给你一排碗，当翻动其中一个时，它和周围两个都翻转，多变元枚举最小值

/*
poj3185
给你20个碗排成一排，当翻动其中一个时，它和周围两个都翻转

*/
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
typedef long double ld;

using namespace std;
const int maxn = 40;

int equ,var;
int a[maxn][maxn];
int b[maxn][maxn];
int x[maxn];
int free_x[maxn];
int free_num;

int Gauss()
{
int max_r,col,k;
free_num = 0;
for(k = 0,col = 0; k < equ && col < var; k++,col++)
{
max_r = k;
for(int i = k+1; i < equ; i++)
{
if(abs(a[i][col]) > abs(a[max_r][col]))
max_r = i;
}
if(a[max_r][col] == 0)
{
k --;
free_x[free_num++] = col;
continue;
}
if(max_r != k)
{
for(int j = col; j < var+1; j++)
swap(a[k][j],a[max_r][j]);

}
for(int i = k + 1; i < equ; i++)
{
if(a[i][col] != 0)
{
for(int j = col; j < var+1; j++)
a[i][j] ^= a[k][j];
}
}

}
for(int i = k; i < equ; i++)
if(a[i][col] != 0)
return -1;
if(k < var) return var-k;

for(int i = var-1; i >= 0; i--)
{
x[i] = a[i][var];
for(int j = i +1; j < var; j++)
x[i] ^= (a[i][j] && x[j]);

}
return 0;

}

int n;
void ini()
{
memset(a,0,sizeof(a));
memset(x,0,sizeof(x));
equ = 20;
var = 20;
for(int i = 0;i < 20;i++)
{
a[i][i] = 1;
if(i > 0) a[i-1][i] = 1;
if(i < 20-1) a[i+1][i]= 1;
}
}

int solve()
{
int t = Gauss();
if(t == -1)
{
return t;
}
else if(t == 0)
{
int ans = 0;
for(int i = 0; i < n*n; i++)
ans += x[i];
return ans;
}
else
{
int ans = 0x3f3f3f3f;
int tot = (1 << t);
for(int i = 0; i < tot; i++)
{
int cnt = 0;
for(int j = 0; j < t; j++)
{
if(i & (1 << j))
{
cnt ++;
x[free_x[j]]= 1;
}
else x[free_x[j]]= 0;
}

for(int j = var-t-1; j >= 0; j--)
{
int dex;
for(dex = j; dex < var; dex++)
if(a[j][dex])
break;
x[dex] = a[j][var];
for(int l = dex +1; l <var ; l++)
{
if(a[j][l])
x[dex] ^= x[l];
}
cnt += x[dex];
}
ans = min(ans,cnt);
}
return ans;
}
}

int main()
{
int tx;
while(scanf("%d",&tx) != EOF)
{
ini();
if(tx == 1)
a[0][20] = 1;
else
a[0][20] = 0;
for(int i=  1; i < 20; i ++)
{
scanf("%d",&tx);
if(tx == 1)
a[i][20] = 1;
else
a[i][20] = 0;
}

int t = solve();
printf("%d\n",t);
}
return 0;
}