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=82;
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=1;i<=n;i++)  //城市初始化
    {
      for(int j=1;j<=n;j++)
      {
             if(i==j)a[i][j]=0;
             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=1;k<=n;k++)
   {
    for(int i=1;i<=n;i++)
    {
     for(int j=1;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=1;i<=m;i++)
    {
      for(int j=1;j<=m;j++)
      {
        a[i][j]=0;  // init()
       for(int k=1;k<=m;k++)
       {
        a[i][j]+=arr[p1][i][k]*arr[p2][k][j];
       }
      }
    }
}

void work()
{
   int t1,t2,t3;
  for(int i=1;i<=n;i++)
  {
      for(int j=1;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=1;t1<=n;t1++)
      {
           //a，b,c三数组不能相同
           if(t1!=i&&t1!=j)
         {
            for( t2=1;t2<=m;t2++)
          {
             for(t3=1;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]=1;
        }
    }
  }
  }
}

int main()
{
   int a,b;
  while(scanf("%d%d",&n,&m)&&n+m!=0)
  {
       for(int i=1;i<=n;i++)
       for(int j=1;j<=m;j++)
         for(int k=1;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
");
        else
            printf("%d
",ans[a][b]);
     }
  }
 return 0;
}