poj3185（开关问题一般解法，以及高斯消元解法）

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/*translation:有20只碗，现在要使得它们全部变成碗口朝上的状态。至少需要多少操作步骤。solution:按照开关问题解，但是要从最左端和最右边分别开始求解，然后比较。note:#：按照poj3276的解法应该就可以解出来，但是为毛在这里还得对从左和从右求解的结果比较？难道poj3276的解法本身就有问题？坑待填...date:2016.11.8*/#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int maxn = 25;int a[maxn], f1[maxn], f2[maxn];int main(){//freopen("in.txt", "r", stdin);while(~scanf("%d", &a[0])){memset(f1, 0, sizeof(f1));memset(f2, 0, sizeof(f2));for(int i = 1; i < 20; i++)scanf("%d", &a[i]);int sum1 = 0, res1 = 0;for(int i = 0; i < 20; i++){//从左if((sum1 + a[i]) & 1){res1++;f1[i] = 1;if(i == 19){res1 = 30;break;}}sum1 += f1[i];if(i - 2 >= 0)sum1 -= f1[i-2];}int sum2 = 0, res2 = 0;for(int i = 19; i >= 0; i--){//从右if((sum2 + a[i]) & 1){res2++;f2[i] = 1;if(i == 0){res2 = 30;break;}}sum2 += f2[i];if(i + 2 < 20)sum2 -= f2[i+2];}printf("%d\n", min(res1, res2));}    return 0;}

/*此版本是高斯消元*/#include <iostream>#include <cstdio>#include <algorithm>#include <cmath>#include <cstring>using namespace std;const int maxn = 300;const int INF = 1e9;int equ, var;int a[maxn][maxn];int x[maxn];int free_x[maxn];int free_num;int gauss(){    int maxr, col, k;    free_num = 0;    for(k = 0, col = 0; k < equ && col < var; k++, col++){        maxr = k;        for(int i = k+1; i < equ; i++){            if(abs(a[i][col]) > abs(a[maxr][col])){                maxr = i;            }        }        if(a[maxr][col] == 0){            k--;            free_x[free_num++] = col;            continue;        }        if(maxr != k){  //交换的到最大主行            for(int i = col; i <= var; i++)                swap(a[k][i], a[maxr][i]);        }        //消元        for(int i = k+1; i < equ; i++){            if(a[i][col] != 0){                for(int j = col; j <= var; j++)                    a[i][j] = a[i][j] ^ a[k][j];            }        }    }    for(int i = k; i < equ; i++)        if(a[k][col])   return -1;    if(k < var)  return var-k;   //返回自由变元的个数    //有唯一解,回代    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 main(){//freopen("in.txt", "r", stdin);    memset(a, 0, sizeof(a));    memset(x, 0, sizeof(x));    memset(free_x, 0, sizeof(free_x));    equ = var = 20;    for(int i = 0; i < 20; i++){a[i][i] = 1;if(i - 1 >= 0)a[i-1][i] = 1;if(i + 1 < 20)a[i+1][i] = 1;    }    for(int i = 0; i < 20; i++)scanf("%d", &a[i][20]);    int tmp = gauss(), ans;    if(tmp == 0){ans = 0;for(int i = 0; i < 20; i++)ans += x[i];printf("%d\n", ans);//有唯一解的时候    }else if(tmp > 0){int total = 1 << tmp, ans = INF;for(int i = 0; i < total; i++){            int cnt = 0;            for(int j = 0; j < tmp; j++){//根据所枚举的子集求出自由变量的值if(i & (1 << j)){x[free_x[j]] = 1;cnt++;}elsex[free_x[j]] = 0;            }            for(int j = var - tmp - 1; j >= 0; j--){//确定自由变量的取值后回代求出剩余变量int id;for(id = j; id < var; id++)if(a[j][id])break;x[id] = a[j][var];for(int k = id+1; k < var; k++){if(a[j][k])x[id] ^= x[k];}cnt += x[id];            }            ans = min(ans, cnt);}printf("%d\n", ans);    }    return 0;}

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