NYOJ 298 点的变换

　　题目链接：298 点的变换

　　这题放在矩阵快速幂里，我一开始想不透它是怎么和矩阵搭上边的，然后写了个暴力的果然超时，上网看了题解后，发现竟然能够构造一些精巧的矩阵来处理，不得不说实在太强大了！ http://blog.****.net/lyhvoyage/article/details/39755595

　　然后我的代码是：

 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<cstdlib>
 #include<iostream>
 using namespace std;
 #define For(i,s,t)    for(int i=s; i<=t; ++i)
 const double pi= acos(-1.0);

 struct matrix{
     int r,c;
     double a[][];
     matrix(): r(),c(){
         memset(a,,sizeof(a));
         a[][]= 1.0;
     }
     matrix(char ch){
         new (this)matrix();
         if(ch=='x' || ch=='X'){
             a[][]= 1.0;
             a[][]= -1.0;
         }
         else if(ch=='y' || ch== 'Y'){
             a[][]= 1.0;
             a[][]= -1.0;
         }
         else if(ch=='E'){
             a[][]= a[][]= 1.0;
         }
     }
     matrix(double p, char ch){
         new (this)matrix();
         if(ch=='s' || ch== 'S'){
             a[][]= a[][]= p;
         }
         else if(ch=='r' || ch=='R'){
             a[][]= a[][]= cos(p);
             a[][]= sin(p);
             a[][]= -sin(p);
         }
     }
     matrix(double dx, double dy, char ch){
         if(ch=='p'|| ch=='P'){
             new (this)matrix();
             r= ;
             a[][]= dx;
             a[][]= dy;
             a[][]= 1.0;
         }
         else if(ch=='M' || ch=='m'){
             new (this)matrix();
             a[][]= a[][]= 1.0;
             a[][]= dx;
             a[][]= dy;
         }
     }
     matrix operator *(const matrix &m2) const {
         matrix mul;
         mul.r= r;
         mul.c= m2.c;
         mul.a[][]= 0.0;
         For(i,,r)    For(j,,m2.c)    For(k,,c)
             mul.a[i][j]+= a[i][k]*m2.a[k][j];
         return mul;
     }
 } point[];

 int main()
 {
     int n,m;
     scanf("%d%d",&n,&m);
     double x,y;
     for(int i=; i<=n; ++i){
         scanf("%lf %lf",&x,&y);
         point[i]= matrix(x,y,'p');
     }
     matrix ans('E');
     while(m--){
         char ch;
         cin>>ch;
         if(ch =='X')    ans= ans*matrix('X');
         else if(ch=='Y')    ans= ans*matrix('Y');
         else if(ch=='M'){
             double dx,dy;
             scanf("%lf %lf",&dx,&dy);
             ans= ans*matrix(dx,dy,'M');
         }
         else if(ch=='S'){
             double p;
             scanf("%lf",&p);
             ans= ans*matrix(p,'S');
         }
         else if(ch=='R'){
             double rad;
             scanf("%lf",&rad);
             rad= rad/*pi;
             ans= ans*matrix(rad,'R');
         }
     }
     for(int i=; i<=n; ++i){
         point[i]= point[i]*ans;
         printf("%.1f %.1f\n",point[i].a[][],point[i].a[][]);
     }
     return ;
 }

　　还是第一次写了超过100行的代码，调试了好久了，唉~~

　　然后，稍微作了下更改，还有另一个版本的：

 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<cstdlib>
 using namespace std;
 #define For(i,s,t)    for(int i=s; i<=t; ++i)
 const double pi= acos(-1.0);

 struct matrix{
     int r,c;
     double a[][];
     void init(int r=, int c=){
         this->r = r;
         this->c = c;
         memset(a,,sizeof(a));
         a[][]= 1.0;
     }
     matrix() {    init();    }
     matrix(char ch){
         init();
         if(ch=='x' || ch=='X'){
             a[][]= 1.0;
             a[][]= -1.0;
         }
         else if(ch=='y' || ch== 'Y'){
             a[][]= 1.0;
             a[][]= -1.0;
         }
         else if(ch=='E'){
             a[][]= a[][]= 1.0;
         }
     }
     matrix(double p, char ch){
         init();
         if(ch=='s' || ch== 'S'){
             a[][]= a[][]= p;
         }
         else if(ch=='r' || ch=='R'){
             a[][]= a[][]= cos(p);
             a[][]= sin(p);
             a[][]= -sin(p);
         }
     }
     matrix(double dx, double dy, char ch){
         if(ch=='p'|| ch=='P'){
             init(,);
             a[][]= dx;
             a[][]= dy;
             a[][]= 1.0;
         }
         else if(ch=='M' || ch=='m'){
             init();
             a[][]= a[][]= 1.0;
             a[][]= dx;
             a[][]= dy;
         }
     }
     matrix operator *(const matrix &m2) const {
         matrix mul;
         mul.init(r,m2.c);
         mul.a[][]= 0.0;
         For(i,,r)    For(j,,m2.c)    For(k,,c)
             mul.a[i][j]+= a[i][k]*m2.a[k][j];
         return mul;
     }
 } point[];

 int main()
 {
     int n,m;
     scanf("%d%d",&n,&m);
     double x,y;
     for(int i=; i<=n; ++i){
         scanf("%lf %lf",&x,&y);
         point[i]= matrix(x,y,'p');
     }
     matrix ans('E');
     while(m--){
         getchar();
         char ch= getchar();
         if(ch =='X')    ans= ans*matrix('X');
         else if(ch=='Y')    ans= ans*matrix('Y');
         else if(ch=='M'){
             double dx,dy;
             scanf("%lf %lf",&dx,&dy);
             ans= ans*matrix(dx,dy,'M');
         }
         else if(ch=='S'){
             double p;
             scanf("%lf",&p);
             ans= ans*matrix(p,'S');
         }
         else if(ch=='R'){
             double rad;
             scanf("%lf",&rad);
             rad= rad/*pi;
             ans= ans*matrix(rad,'R');
         }
     }
     for(int i=; i<=n; ++i){
         point[i]= point[i]*ans;
         printf("%.1f %.1f\n",point[i].a[][],point[i].a[][]);
     }
     return ;
 }