ABC189 E-Rotate and Flip (coordinate transformation, matrix multiplication)

Title:

solution:

1.顺时针旋转90度:
(x,y)->(y,-x)

2.逆时针旋转90度:
(x,y)->(-y,x)

3.关于x=p对称:
(x,y)->(2p-x,y)

4.关于y=p对称:
(x,y)->(x,2p-y)

可以推出以下矩阵,维护以下矩阵前缀积即可.

code：

#include <bits/stdc++.h>
#define int long long
#define PI pair<int,int>
using namespace std;
const int maxm=2e6+5;
struct Mat{
    
    
    int a[3][3];
    void init(){
    
    
        memset(a,0,sizeof a);
    }
    Mat operator*(const Mat& x){
    
    
        Mat ans;ans.init();
        for(int k=0;k<3;k++){
    
    
            for(int i=0;i<3;i++){
    
    
                for(int j=0;j<3;j++){
    
    
                    ans.a[i][j]+=a[i][k]*x.a[k][j];
                }
            }
        }
        return ans;
    }
};
int x[maxm],y[maxm];
Mat T[maxm];
int n,m,q;
void solve(){
    
    
    cin>>n;
    for(int i=1;i<=n;i++)cin>>x[i]>>y[i];
    cin>>m;
    T[0]={
    
    1,0,0,0,1,0,0,0,1};//T[0]是单位矩阵
    for(int i=1;i<=m;i++){
    
    
        int op;cin>>op;
        Mat t;
        if(op==1){
    
    //顺时针旋转90度
            t={
    
    0,1,0,-1,0,0,0,0,1};
        }else if(op==2){
    
    //逆时针旋转90度
            t={
    
    0,-1,0,1,0,0,0,0,1};
        }else if(op==3){
    
    //关于x=p对称
            int p;cin>>p;
            t={
    
    -1,0,2*p,0,1,0,0,0,1};
        }else if(op==4){
    
    //关于y=p对称
            int p;cin>>p;
            t={
    
    1,0,0,0,-1,2*p,0,0,1};
        }
        T[i]=t*T[i-1];
    }
    cin>>q;
    for(int i=1;i<=q;i++){
    
    
        int a,b;cin>>a>>b;
        Mat t={
    
    x[b],0,0,y[b],0,0,1,0,0};
        t=T[a]*t;
        cout<<t.a[0][0]<<' '<<t.a[1][0]<<endl;
    }
}
signed main(){
    
    
    ios::sync_with_stdio(0);
    solve();
    return 0;
}

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