hdu 2807 The Shortest Path(矩阵+floyd)

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

 

此题要先找出哪些城市可以连边，可以通过矩阵相乘来实现，得出的矩阵与第三个矩阵比较，看看是否可以连边。

算出所有edge后，再通过floyd来算出任意两个城市的最短距离。

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<set>
using namespace std;
#define inf 1<<26
#define N 86
int n,m;
int edge[N][N];
struct Matrix
{
    int mp[N][N];
}matrix[N];
Matrix Mul(Matrix a,Matrix b)
{
    Matrix res;
    for(int i=1;i<=m;i++)
    {
        for(int j=1;j<=m;j++)
        {
            res.mp[i][j]=0;
            for(int k=1;k<=m;k++)
            {
                res.mp[i][j]=res.mp[i][j]+(a.mp[i][k]*b.mp[k][j]);
            }
        }
    }
    return res;
}
bool judge(Matrix a,Matrix b)
{
    for(int i=1;i<=m;i++)
    {
        for(int j=1;j<=m;j++)
        {
            if(a.mp[i][j]!=b.mp[i][j])
              return false;
        }
    }
    return true;
}
void init()
{
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
            if(i==j)
                 edge[i][j]=0;
            else 
            {
                  edge[i][j]=inf;
                }
        }
    }
}
void floyd()
{
    for(int k=1;k<=n;k++)
    {
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(edge[i][j]>edge[i][k]+edge[k][j])
                  edge[i][j]=edge[i][k]+edge[k][j];
            }
        }
    }
}
int main()
{
    while(scanf("%d%d",&n,&m)==2 && n+m)
    {
        //memset(edge,0,sizeof(edge));
        
        init();
        
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                for(int k=1;k<=m;k++)
                {
                    scanf("%d",&matrix[i].mp[j][k]);
                }
            }
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(i==j)
                  continue;
                Matrix tmp = Mul(matrix[i],matrix[j]);
                for(int k=1;k<=n;k++)
                {
                    if(i==k || j==k)
                      continue;
                      if(judge(tmp,matrix[k]))
                        edge[i][k]=1;
                   }
            }
        }
        
        floyd();
        
        int c;
        scanf("%d",&c);
        while(c--)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            if(edge[u][v]!=inf)
              printf("%d
",edge[u][v]);
            else 
              printf("Sorry
");
        }
    }
    return 0;
}