高斯消元\高斯约当 模板

实数版：O（n^3)

解方程：

#include <cstdio>#include <algorithm>#define rep(i,l,r) for (int i=l;i<=r;++i)const double EPS=1e-8;int sign(const double &x){return x<-EPS?-1:x>EPS;}const int MAX_N=105,MAX_M=1005;double a[MAX_N][MAX_M+MAX_N];int n,m;void gauss(){    rep(i,1,n){        int p=1;        rep(j,i,n) if (sign(a[j][i])){p=j;break;} //这里应该写成绝对值最大的精度高一些        rep(k,1,n+m) std::swap(a[i][k],a[p][k]);        rep(j,1,n) if (i!=j){            double r=a[j][i]/a[i][i];            rep(k,i,n+m) a[j][k]-=r*a[i][k];            }        }    rep(i,n+1,n+m) rep(j,1,n)        printf("%.3f%c",a[j][i]/a[j][j],j==n?'\n':' ');}int main(){    scanf("%d%d",&n,&m);    rep(i,1,n) rep(j,1,n) scanf("%lf",&a[i][j]);    rep(i,n+1,n+m) rep(j,1,n) scanf("%lf",&a[j][i]);    gauss();}

整数版：

有O(N^3)   （LCM）

   O(N^3logM)  （模拟GCD）

O(N^3)

(C一份~~~)

int Gauss()  {      int i,j,k;      int max_r;      int ta,tb;      int free_x_num;      int free_index;      int LCM,tmp;      int col = 0;      for(k=0; k<equ && col<var; k++,col++)      {          max_r = k;          for(i=k+1; i<equ; i++)              if(abs(a[i][col]) > abs(a[max_r][col])) max_r = i;          if(max_r != k)              for(j=k; j<var+1; j++)                  swap(a[k][j],a[max_r][j]);          if(a[k][col] == 0)          {              k--;              continue;          }          for(i=k+1; i<equ; i++)          {              if(a[i][col] != 0)              {                  LCM = lcm(abs(a[i][col]),abs(a[k][col]));                  ta = LCM / abs(a[i][col]);                  tb = LCM / abs(a[k][col]);                  if(a[i][col] * a[k][col] < 0) tb = - tb;                  for(j = col ; j<var+1; j++)                      a[i][j] = a[i][j]*ta - a[k][j]*tb;              }          }      }  

XOR方程组（没有压位）：O(N^3)

压位：O(NlogMAX)

#include <cstdio>#include <algorithm>#define rep(i,l,r) for (int i=l;i<=r;++i)#define per(i,r,l) for (int i=r;i>=l;--i)const int MAX_N=40;bool a[MAX_N][MAX_N];int n,m;void gauss(){rep(i,1,n){rep(j,i,n) if (a[j][i]>a[i][i]) rep(k,1,n+1) std::swap(a[i][k],a[j][k]);rep(j,1,n) if (i!=j&&a[j][i])   rep(k,i,n+1) a[j][k]^=a[i][k];}}int ans=100;bool x[MAX_N];void dfs(int v,int tot){  //自由元枚举if (tot>ans) return;if (v<=0){ans=tot;return;}if (a[v][v]){x[v]=a[v][n+1];rep(k,v+1,n) x[v]^=a[v][k]&x[k];dfs(v-1,tot+x[v]);}else{x[v]=1;dfs(v-1,tot+1);x[v]=0;dfs(v-1,tot);}}int main(){freopen("lights.in","r",stdin);freopen("lights.out","w",stdout);scanf("%d%d",&n,&m);rep(i,1,n) a[i][i]=a[i][n+1]=1;rep(i,1,m){int x,y;scanf("%d%d",&x,&y);a[x][y]=a[y][x]=1;}gauss();dfs(n,0);printf("%d\n",ans);}

ORZ MATO：http://www.cppblog.com/MatoNo1/archive/2012/05/20/175404.html

AND ：http://wenku.baidu.com/link?url=PsKxF1uwSUIoTx9zuOpTQbSPuiQrTtDSx_TnGXI6r_nw1Evg37lqXOJL3CODAokkV9JVhw4qqXMGhyItwihn-cjan1jKth7-PQoB0s9nN1K