POJ3318 Matrix Multiplication 

  1 /*
  2  POJ3318 Matrix Multiplication
  3  http://poj.org/problem?id=3318
  4  随机化算法 矩阵
  5  随机一个1×n的矩阵h
  6  在ab=c左右两边左乘h
  7  验证是否相等
  8  然而不知道为什么g++交就会RE，
  9  c++交AC
 10  *
 11  */
 12 #include <cstdio>
 13 #include <algorithm>
 14 #include <cstring>
 15 #include <cmath>
 16 #include <vector>
 17 #include <queue>
 18 #include <iostream>
 19 #include <map>
 20 #include <set>
 21 #include <ctime>
 22 #include <cstdlib>
 23 //#define test
 24 using namespace std;
 25 const int Nmax=505;
 26 const int mod=500;
 27 int n;
 28 struct Matrix
 29 {
 30     int n,m;
 31     int num[Nmax][Nmax];
 32     Matrix()
 33     {
 34         n=0;
 35         m=0;
 36     }
 37     Matrix &operator = (const Matrix b)
 38     {
 39         n=b.n;
 40         m=b.m;
 41         for(int i=1;i<=n;i++)
 42             for(int j=1;j<=m;j++)
 43                 num[i][j]=b.num[i][j];
 44         //show();
 45         return *this;
 46     }
 47     //Matrix(Matrix &a)
 48     //{
 49         //*this=a;
 50     //}
 51     Matrix(int _n,int _m,int **_num)
 52     {
 53         n=_n;
 54         m=_m;
 55         for(int i=1;i<=n;i++)
 56             for(int j=1;j<=m;j++)
 57                 num[i][j]=_num[i][j];
 58     }
 59     friend Matrix operator + (Matrix a,Matrix b)
 60     {
 61         Matrix ans;
 62         ans.n=a.n;
 63         ans.m=a.m;
 64         for(int i=1;i<=ans.n;i++)
 65             for(int j=1;j<=ans.m;j++)
 66                 ans.num[i][j]=a.num[i][j]+b.num[i][j];
 67         return ans;
 68     }
 69     friend Matrix operator * (Matrix a,Matrix b)
 70     {
 71         Matrix ans;
 72         if(a.m!=b.n)
 73         {
 74             printf("error!\n");
 75             return ans;
 76         }
 77         ans.n=a.n,ans.m=b.m;
 78         int tmp=a.m;
 79         for(int i=1;i<=ans.n;i++)
 80             for(int j=1;j<=ans.m;j++)
 81                 ans.num[i][j]=0;
 82         for(int i=1;i<=ans.n;i++)
 83             for(int j=1;j<=ans.m;j++ )
 84                 for(int k=1;k<=tmp;k++)
 85                     ans.num[i][j]+=a.num[i][k]*b.num[k][j];
 86         //ans.show();
 87         return ans;
 88     }
 89     friend bool operator != (Matrix a,Matrix b)
 90     {
 91         if(a.n!=b.n || a.m!=b.m)
 92             return 1;
 93         for(int i=1;i<=a.n;i++)
 94             for(int j=1;j<=a.m;j++)
 95                 if(a.num[i][j]!=b.num[i][j])
 96                     return 1;
 97         return 0;
 98     }
 99     void show()
100     {
101         printf("n:%d m:%d\n",n,m);
102         for(int i=1;i<=n;i++)
103             for(int j=1;j<=m;j++)
104                 printf("%d%c",num[i][j],j==m? '\n':' ');
105     }
106 };
107 Matrix a,b,c,h;
108 //Matrix tmpa,tmpb,tmpc;
109 int work()
110 {
111     srand(unsigned(time(0)));
112     h.n=1,h.m=n;
113     for(int t=1;t<=5;t++)
114     {
115         for(int i=1;i<=n;i++)
116         {
117             h.num[1][i]=rand()%mod;
118 
119         }
120         //h.show();
121         //tmpa=h*a;
122         //tmpb=tmpa*b;
123         //tmpc=h*c;
124         //tmpa.show();
125         //tmpb.show();
126         //tmpc.show();
127         if( ( (h*a)*b )!=(h*c) )
128             return 0;
129     }
130 
131     return 1;
132 }
133 int main()
134 {
135     #ifdef test
136     #endif
137     //freopen("poj3318.in","r",stdin);
138     scanf("%d",&n);
139     a.n=a.m=b.n=b.m=c.n=c.m=n;
140     for(int i=1;i<=n;i++)
141         for(int j=1;j<=n;j++)
142             scanf("%d",&a.num[i][j]);
143     for(int i=1;i<=n;i++)
144         for(int j=1;j<=n;j++)
145             scanf("%d",&b.num[i][j]);
146     for(int i=1;i<=n;i++)
147         for(int j=1;j<=n;j++)
148             scanf("%d",&c.num[i][j]);
149     if(work())
150         printf("YES\n");
151     else
152         printf("NO\n");
153     return 0;
154 }

 

posted @ 2017-04-06 13:49  BBBob  阅读(216)  评论(0)    收藏  举报