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：

 

  1 #include<cstdio>

  2 #include<cstring>

  3 #define inf 0x3f3f3f3f

  4 using namespace std;

  5 

  6 const int maxn=82;

  7 int arr[maxn][maxn][maxn];

  8 int ans[maxn][maxn];

  9 int tem[maxn][maxn];

 10 

 11 int n,m,w;

 12 

 13 void init(int a[][maxn])

 14 {

 15     for(int i=1;i<=n;i++)  //城市初始化

 16     {

 17       for(int j=1;j<=n;j++)

 18       {

 19              if(i==j)a[i][j]=0;

 20              else a[i][j]=inf;

 21       }

 22     }

 23 }

 24 

 25 void floyd(int a[][maxn])    //运用floyd算法求城市间的最短路径

 26 {

 27    for(int k=1;k<=n;k++)

 28    {

 29     for(int i=1;i<=n;i++)

 30     {

 31      for(int j=1;j<=n;j++)

 32      {

 33         if(ans[i][j]>ans[i][k]+ans[k][j])

 34           ans[i][j]=ans[i][k]+ans[k][j];

 35      }

 36     }

 37    }

 38 }

 39 

 40 void Matrix(int a[][maxn],int p1,int p2)

 41 {

 42     for(int i=1;i<=m;i++)

 43     {

 44       for(int j=1;j<=m;j++)

 45       {

 46         a[i][j]=0;  // init()

 47        for(int k=1;k<=m;k++)

 48        {

 49         a[i][j]+=arr[p1][i][k]*arr[p2][k][j];

 50        }

 51       }

 52     }

 53 }

 54 

 55 void work()

 56 {

 57    int t1,t2,t3;

 58   for(int i=1;i<=n;i++)

 59   {

 60       for(int j=1;j<=n;j++)

 61     {

 62        if(i==j) continue; //a，b 两数组不能相同

 63          Matrix(tem,i,j);  //两个矩阵相乘

 64       for(t1=1;t1<=n;t1++)

 65       {

 66            //a，b,c三数组不能相同

 67            if(t1!=i&&t1!=j)

 68          {

 69             for( t2=1;t2<=m;t2++)

 70           {

 71              for(t3=1;t3<=m;t3++)

 72             {

 73               //得到的结果相比较

 74               if(tem[t2][t3]!=arr[t1][t2][t3])

 75                 goto   loop;

 76             }

 77           }

 78            loop:

 79                if(t3>m)

 80                   ans[i][t1]=1;

 81         }

 82     }

 83   }

 84   }

 85 }

 86 

 87 int main()

 88 {

 89    int a,b;

 90   while(scanf("%d%d",&n,&m)&&n+m!=0)

 91   {

 92        for(int i=1;i<=n;i++)

 93        for(int j=1;j<=m;j++)

 94          for(int k=1;k<=m;k++)

 95              scanf("%d",&arr[i][j][k]);

 96       init(ans);

 97       work();

 98      floyd(ans);

 99      scanf("%d",&w);

100      while(w--)

101      {

102         scanf("%d%d",&a,&b);

103         if(ans[a][b]==inf)

104             printf("Sorry\n");

105         else

106             printf("%d\n",ans[a][b]);

107      }

108   }

109  return 0;

110 }

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