学习笔记--矩阵相关
高斯消元
找出变量间关系
- 高斯-约当消元法
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <cmath>
using namespace std;
const int maxn=103;
const double eps=1e-8;
int n;
double a[maxn][maxn],b[maxn];//a--系数 b--常数
inline bool gs(){
double rate;
for(register int i=1;i<=n;i++){//消元
int k=i;
for(register int j=i+1;j<=n;j++)
if(fabs(a[j][i])>fabs(a[k][i]))k=j;//选取最大做主元防止误差
if(fabs(rate=a[k][i])<eps)return 0;//主元无系数 无解
for(register int j=i;j<=n;j++)swap(a[i][j],a[k][j]);//将主元那一行换到第i行来
swap(b[i],b[k]);
for(register int j=i;j<=n;j++)a[i][j]=a[i][j]/rate;//将主元系数化为1
b[i]=b[i]/rate;
for(register int k=1;k<=n;k++){
if(k==i)continue;
rate=a[k][i];
for(int j=i;j<=n;j++)a[k][j]=a[k][j]-a[i][j]*rate;
b[k]=b[k]-b[i]*rate;
}
}
return 1;
}
int main(){
scanf("%d",&n);
for(register int i=1;i<=n;i++){
for(register int j=1;j<=n;j++){
scanf("%lf",&a[i][j]);
}
scanf("%lf",&b[i]);
}
if(gs()==1){
for(register int i=1;i<=n;i++)printf("%.2lf\n",b[i]);
}
else{
puts("No Solution");
}
return 0;
}
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <cctype>
using namespace std;
const int maxn=13;
const double eps=1e-8;
double a[maxn][maxn],b[maxn][maxn],c[maxn];
int n;
inline void gs(){
double rate;
int mx;
for(register int i=1;i<=n;i++){
mx=i;
for(register int j=i+1;j<=n;j++){
if(fabs(b[j][i])>fabs(b[mx][i]))mx=j;
}
//if(fabs(rate=b[mx][i])<eps)return ;
rate=b[mx][i];
for(register int j=i;j<=n;j++){
swap(b[i][j],b[mx][j]);
}
swap(c[i],c[mx]);
for(register int j=i;j<=n;j++)b[i][j]=b[i][j]/rate;
c[i]=c[i]/rate;
for(register int k=1;k<=n;k++){
if(k==i)continue;
rate=b[k][i];
for(register int j=i;j<=n;j++)b[k][j]=b[k][j]-b[i][j]*rate;
c[k]=c[k]-c[i]*rate;
}
}
return ;
}
int main(){
scanf("%d",&n);
for(register int i=1;i<=n+1;i++){
for(register int j=1;j<=n;j++){
scanf("%lf",&a[i][j]);
}
}
for(register int i=1;i<=n;i++){
for(register int j=1;j<=n;j++){
b[i][j]=2*(a[i][j]-a[i+1][j]);
c[i]+=a[i][j]*a[i][j]-a[i+1][j]*a[i+1][j];
}
}
gs();
for(register int i=1;i<=n;i++){
printf("%.3lf ",c[i]);
}
return 0;
}
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期望&马尔可夫过程&&UVa 10828
待填坑
-
异或
待填坑
线性基
- 待填坑
矩阵加速递推
关键构造友矩阵
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cctype>
#define ll long long
using namespace std;
const int maxn=107;
const long long p=1000000007;
struct Matrix{
int n,m;
ll ma[maxn][maxn];
Matrix(int rk){n=m=rk;for(register int i=1;i<=rk;i++)ma[i][i]=1;}
Matrix(int rk_n,int rk_m){n=rk_n,m=rk_m;}
Matrix(){;}
Matrix operator *(const Matrix &x)const{
Matrix c=Matrix(n,x.m);
for(register int i=1;i<=n;i++){
for(register int j=1;j<=x.m;j++){
ll tmp=0;
for(register int k=1;k<=m;k++){
tmp+=ma[i][k]*x.ma[k][j];
if(tmp>19260817)tmp=tmp%p;
}
c.ma[i][j]=tmp;
}
}
return c;
}
Matrix operator ^(const int &C)const{
int c=C;
Matrix ans=Matrix(n);
Matrix res=*this;
while(c){
if(c&1){ans=ans*res;}
res=res*res;
c=c>>1;
}
return ans;
}
Matrix operator ^(const ll &C)const{
ll c=C;
Matrix ans=Matrix(n);
Matrix res=*this;
while(c){
if(c&1){ans=ans*res;}
res=res*res;
c=c>>1;
}
return ans;
}
};
int main(){
ll n;
Matrix f0(2,1);f0.ma[1][1]=1,f0.ma[2][1]=1;
Matrix A(2);A.ma[1][1]=1,A.ma[1][2]=1,A.ma[2][1]=1,A.ma[2][2]=0;
scanf("%lld",&n);
if(n<=2){
cout<<1<<endl;
return 0;
}
//Matrix ans=mul(Matrix_ksm(A,n),f0);
//Matrix ans=(A^(n-1))*f0;
A=A^(n-1);
Matrix ans=f0*A;
printf("%lld\n",ans.ma[1][1]);
return 0;
}
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注意！！！矩阵定义中一定要加memset，否则可能有非0元素，太恶心了
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <cctype>
#include <cmath>
#define ll long long
const ll p=1000000007;
const int maxn=15;
int t;
struct Matrix{
int n,m;
ll ma[maxn][maxn];//注意！！！！一定要memset
Matrix(int _n){n=m=_n;memset(ma,0,sizeof(ma));for(register int i=1;i<=n;i++)ma[i][i]=1;}
Matrix(int _n,int _m){n=_n,m=_m;}
Matrix(){;}
Matrix operator *(const Matrix& b)const {
Matrix c=Matrix(n,b.m);
for(register int i=1;i<=n;i++){
for(register int j=1;j<=b.m;j++){
ll tmp=0;
for(register int k=1;k<=m;k++){
tmp+=ma[i][k]*b.ma[k][j];
if(tmp>19260817)tmp=tmp%p;
}
c.ma[i][j]=tmp;
}
}
return c;
}
Matrix operator ^(const int &C)const {
int c=C;
Matrix ans=Matrix(n);
Matrix res=*this;
while(c){
if(c&1)ans=ans*res;
res=res*res;
c=c>>1;
}
return ans;
}
Matrix operator ^(const ll &C)const {
ll c=C;
Matrix ans=Matrix(n);
/*for(int i=1;i<=3;i++){
for(int j=1;j<=3;j++)printf("%lld ",ans.ma[i][j]);
puts("");
}*/
Matrix res=*this;
while(c){
if(c&1)ans=ans*res;
res=res*res;
c=c>>1;
}
return ans;
}
};
int main(){
ll n;
scanf("%d",&t);
while(t--){
scanf("%lld",&n);
if(n<=3){puts("1");continue;}
Matrix A=Matrix(3,3);
A.ma[1][1]=1;A.ma[1][2]=0;A.ma[1][3]=1;
A.ma[2][1]=1;A.ma[2][2]=0;A.ma[2][3]=0;
A.ma[3][1]=0;A.ma[3][2]=1;A.ma[3][3]=0;
Matrix f0=Matrix(3,1);
f0.ma[1][1]=1,f0.ma[2][1]=1,f0.ma[3][1]=1;
A=A^(n-1);
//std::cout<<'*'<<A.ma[1][1]<<std::endl;
/*for(int i=1;i<=3;i++){
for(int j=1;j<=3;j++)printf("%lld ",A.ma[i][j]);
puts("");
}*/
f0=f0*A;
printf("%lld\n",f0.ma[1][1]);
}
return 0;
}
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进阶
https://www.luogu.org/problemnew/show/P1707 构造转移矩阵比较麻烦还要快速乘
我的题解：https://www.cnblogs.com/Rye-Catcher/p/9089459.html
https://www.luogu.org/problemnew/show/P2643
待填坑
矩阵树定理（Matrix-Tree）
-
待填坑
推荐阅读：http://www.cnblogs.com/zj75211/p/8039443.html
完全二分图生成树的个数是
\(n^{m−1}*m^{n−1}\)bzoj2467 bzoj1002 SP3381
