poj 3185 The Water Bowls 【高斯消元 + 枚举自由变元】

The Water Bowls

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5085 Accepted: 1991

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：有20个为0或1的整数，已知改变i位置的数字时，会同时改变i-1和i+1位置的数字，改变效应——0->1 或者 1->0。问你最少的操作使得20个整数全为0。

思路：20个方程，20个变元——高斯消元 + 妆压枚举自由变元 就可以了。

AC代码：

#include <cstdio>#include <cstring>#include <algorithm>#define MAXN 25#define INF 0x3f3f3f3fusing namespace std;int a[MAXN][MAXN];int free_rec[MAXN];int x[MAXN];int equ, var;void init_a(){    equ = var = 20;    memset(a, 0, sizeof(a));    for(int i = 1; i < 20; i++)    {        scanf("%d", &a[i][var]), a[i][var] ^= 0;        a[i][i] = 1;        a[i][i-1] = 1;        if(i < 19)            a[i][i+1] = 1;    }}int Gauss(){    int max_r, k;    int col = 0;    int num = 0;    for(k = 0; k < equ && col < var; k++, col++)    {        max_r = k;        for(int i = k+1; i < equ; i++)            if(a[i][col] > a[max_r][i])                max_r = i;        if(max_r != k)            for(int i = col; i < var+1; i++)                swap(a[k][i], a[max_r][i]);        if(a[k][col] == 0)        {            k--;            free_rec[num++] = col;            continue;        }        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+1; i < equ; i++)        if(a[i][col] != 0)            return -1;    if(var > k)        return var - k;    return 0;}void solve(int S){    int state = (1<<S);    int ans = INF;    for(int i = 0; i < state; i++)    {        int cnt = 0;        for(int j = 0; j < S; j++)        {            if((1<<j) & i)            {                cnt++;                x[free_rec[j]] = 1;            }            else                x[free_rec[j]] = 0;        }        for(int j = var-S-1; j >= 0; j--)        {            int temp = a[j][var];            for(int l = j+1; l < var; l++)                if(a[j][l])                    temp ^= x[l];            x[j] = temp;            cnt += x[j] ? 1 : 0;        }        ans = min(ans, cnt);    }    printf("%d\n", ans);}int main(){    while(scanf("%d", &a[0][20]) != EOF)    {        a[0][0] = 1, a[0][1] = 1, a[0][20] ^= 0;        init_a();        solve(Gauss());    }    return 0;}