邝斌的ACM模板（高斯消元法求方程组的解）

本博客整理自邝斌的ACM模板

2.10、高斯消元法求方程组的解

2.10.1 一类开关问题，对 2 取模的 01 方程组

POJ 1681 需要枚举自由变元，找解中 1 个数最少的

//对2取模的01方程组
const int MAXN = 300;
//有equ个方程，var个变元。增广矩阵行数为equ,列数为var+1,分别为0到var
int equ,var;
int a[MAXN][MAXN]; //增广矩阵
int x[MAXN]; //解集
int free_x[MAXN];//用来存储自由变元（多解枚举自由变元可以使用）
int free_num;//自由变元的个数

//返回值为-1表示无解，为0是唯一解，否则返回自由变元个数
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;//这个是自由变元
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; //无解
if(k < var) return var-k;//自由变元个数
//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 n;
void init()
{
memset(a,0,sizeof(a));
memset(x,0,sizeof(x));
equ = n*n;
var = n*n;
for(int i = 0; i < n; i++)
for(int j = 0; j < n; j++)
{
int t = i*n+j;
a[t][t] = 1;
if(i > 0)a[(i-1)*n+j][t] = 1;
if(i < n-1)a[(i+1)*n+j][t] = 1;
if(j > 0)a[i*n+j-1][t] = 1;
if(j < n-1)a[i*n+j+1][t] = 1;
}
}
void solve()
{
int t = Gauss();
if(t == -1)
{
printf("inf\n");
return;
}
else if(t == 0)
{
int ans = 0;
for(int i = 0; i < n*n; i++)    ans += x[i];
printf("%d\n",ans);
return;
}
else
{
//枚举自由变元
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))
{
x[free_x[j]] = 1;
cnt++;
}
else x[free_x[j]] = 0;
}
for(int j = var-t-1; j >= 0; j--)
{
int idx;
for(idx = j; idx < var; idx++)
if(a[j][idx])
break;
x[idx] = a[j][var];
for(int l = idx+1; l < var; l++)
if(a[j][l])

4000
x[idx] ^= x[l];
cnt += x[idx];
}
ans = min(ans,cnt);
}
printf("%d\n",ans);
}
}
char str[30][30];
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
init();
for(int i = 0; i < n; i++)
{
scanf("%s",str[i]);
for(int j = 0; j < n; j++)
{
if(str[i][j] == 'y')
a[i*n+j][n*n] = 0;
else a[i*n+j][n*n] = 1;
}
}
solve();
}
return 0;
}

2.10.2 解同余方程组

POJ 2947 Widget Factory

//求解对MOD取模的方程组
const int MOD = 7;
const int MAXN = 400;
int a[MAXN][MAXN];//增广矩阵
int x[MAXN];//最后得到的解集

inline int gcd(int a,int b)
{
while(b != 0)
{
int t = b;
b = a%b;
a = t;
}
return a;
}
inline int lcm(int a,int b)
{
return a/gcd(a,b)*b;
}
long long inv(long long a,long long m)
{
if(a == 1)return 1;
return inv(m%a,m)*(m-m/a)%m;
}
int Gauss(int equ,int var)
{
int max_r,col,k;
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--;
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)
{
int LCM = lcm(abs(a[i][col]),abs(a[k][col]));
int ta = LCM/abs(a[i][col]);
int tb = LCM/abs(a[k][col]);
if(a[i][col]*a[k][col] < 0)tb = -tb;
for(int j = col; j < var+1; j++)
a[i][j] = ((a[i][j]*ta - a[k][j]*tb)%MOD + MOD)%MOD;
}
}
}
for(int i = k; i < equ; i++)
if(a[i][col] != 0)
return -1; //无解
if(k < var) return var-k;//多解
for(int i = var-1; i >= 0; i--)
{
int temp = a[i][var];
for(int j = i+1; j < var; j++)
{
if(a[i][j] != 0)
{
temp -= a[i][j]*x[j];
temp = (temp%MOD + MOD)%MOD;
}
}
x[i] = (temp*inv(a[i][i],MOD))%MOD;
}
return 0;
}
int change(char s[])
{
if(strcmp(s,"MON") == 0) return 1;
else if(strcmp(s,"TUE")==0) return 2;
else if(strcmp(s,"WED")==0) return 3;
else if(strcmp(s,"THU")==0) return 4;
else if(strcmp(s,"FRI")==0) return 5;
else if(strcmp(s,"SAT")==0) return 6;
else return 7;
}
int main()
{
int n,m;
while(scanf("%d%d",&n,&m) == 2)
{
if(n == 0 && m == 0)break;
memset(a,0,sizeof(a));
char str1[10],str2[10];
int k;
for(int i = 0; i < m; i++)
{
scanf("%d%s%s",&k,str1,str2);
a[i]
= ((change(str2) - change(str1) + 1)%MOD + MOD)%MOD;
int t;
while(k--)
{
scanf("%d",&t);
t--;
a[i][t] ++;
a[i][t]%=MOD;
}
}
int ans = Gauss(m,n);
if(ans == 0)
{
for(int i = 0; i < n; i++)
if(x[i] <= 2)
x[i] += 7;
for(int i = 0; i < n-1; i++)printf("%d ",x[i]);
printf("%d\n",x[n-1]);
}
else if(ans == -1)printf("Inconsistent data.\n");
else printf("Multiple solutions.\n");
}
return 0;
}