hdu-----(2807)The Shortest Path(矩阵+Floyd)

The Shortest Path

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2440    Accepted Submission(s): 784

Problem Description

There
are N cities in the country. Each city is represent by a matrix size of
M*M. If city A, B and C satisfy that A*B = C, we say that there is a
road from A to C with distance 1 (but that does not means there is a
road from C to A).
Now the king of the country wants to ask me some problems, in the format:
Is there is a road from city X to Y?
I have to answer the questions quickly, can you help me?

 

Input

Each
test case contains a single integer N, M, indicating the number of
cities in the country and the size of each city. The next following N
blocks each block stands for a matrix size of M*M. Then a integer K
means the number of questions the king will ask, the following K lines
each contains two integers X, Y(1-based).The input is terminated by a
set starting with N = M = 0. All integers are in the range [0, 80].

 

Output

For
each test case, you should output one line for each question the king
asked, if there is a road from city X to Y? Output the shortest distance
from X to Y. If not, output "Sorry".

 

Sample Input

3 2
1 1
2 2
1 1
1 1
2 2
4 4
1
1 3
3 2
1 1
2 2
1 1
1 1
2 2
4 3
1
1 3
0 0

 

Sample Output

1
Sorry

 

Source

HDU 2009-4 Programming Contest

 

 

题目很清楚： 

如果满足A*B=C 那么就说A到C是联通的....

反之则为不连通...

这里需要用到的求最短路径，对于稠密图的，用弗洛伊德算法比狄斯喹诺算法要好一点。

至于要优化，看来结题报告之后，觉得还是不大靠谱，就没有写，写的是一个朴素的矩阵相乘算法+floyd算法

代码：

 #include<cstdio>
 #include<cstring>
 #define inf 0x3f3f3f3f
 using namespace std;

 const int maxn=;
 int arr[maxn][maxn][maxn];
 int ans[maxn][maxn];
 int tem[maxn][maxn];

 int n,m,w;

 void init(int a[][maxn])
 {
     for(int i=;i<=n;i++)  //城市初始化
     {
       for(int j=;j<=n;j++)
       {
              if(i==j)a[i][j]=;
              else a[i][j]=inf;
       }
     }
 }

 void floyd(int a[][maxn])    //运用floyd算法求城市间的最短路径
 {
    for(int k=;k<=n;k++)
    {
     for(int i=;i<=n;i++)
     {
      for(int j=;j<=n;j++)
      {
         if(ans[i][j]>ans[i][k]+ans[k][j])
           ans[i][j]=ans[i][k]+ans[k][j];
      }
     }
    }
 }

 void Matrix(int a[][maxn],int p1,int p2)
 {
     for(int i=;i<=m;i++)
     {
       for(int j=;j<=m;j++)
       {
         a[i][j]=;  // init()
        for(int k=;k<=m;k++)
        {
         a[i][j]+=arr[p1][i][k]*arr[p2][k][j];
        }
       }
     }
 }

 void work()
 {
    int t1,t2,t3;
   for(int i=;i<=n;i++)
   {
       for(int j=;j<=n;j++)
     {
        if(i==j) continue; //a，b 两数组不能相同
          Matrix(tem,i,j);  //两个矩阵相乘
       for(t1=;t1<=n;t1++)
       {
            //a，b,c三数组不能相同
            if(t1!=i&&t1!=j)
          {
             for( t2=;t2<=m;t2++)
           {
              for(t3=;t3<=m;t3++)
             {
               //得到的结果相比较
               if(tem[t2][t3]!=arr[t1][t2][t3])
                 goto   loop;
             }
           }
            loop:
                if(t3>m)
                   ans[i][t1]=;
         }
     }
   }
   }
 }

 int main()
 {
    int a,b;
   while(scanf("%d%d",&n,&m)&&n+m!=)
   {
        for(int i=;i<=n;i++)
        for(int j=;j<=m;j++)
          for(int k=;k<=m;k++)
              scanf("%d",&arr[i][j][k]);
       init(ans);
       work();
      floyd(ans);
      scanf("%d",&w);
      while(w--)
      {
         scanf("%d%d",&a,&b);
         if(ans[a][b]==inf)
             printf("Sorry\n");
         else
             printf("%d\n",ans[a][b]);
      }
   }
  return ;
 }