HDU2807The Shortest Path（SPFA或Floyd）

The Shortest Path

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2648    Accepted Submission(s): 835

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 

<pre name="code" class="cpp">#include<stdio.h>
#include<queue>
#include<string.h>
using namespace std;
const int N = 100;
const int inf = 9999999;

struct matrix
{
    int m[N][N];
};

int n,m,map[N][N],dis[N][N];
matrix M[N];

void init()
{
    for(int i=0;i<n;i++)
    for(int j=0;j<n;j++)
    map[i][j]=0,dis[i][j]=inf;
}
matrix multiply(const matrix &x,const matrix &y)
{
    matrix temp;
    memset(temp.m,0,sizeof(temp.m));
    for(int i=0; i<m; i++)
    {
        for(int j=0; j<m; j++)
        {
            if(x.m[i][j]==0) continue;
            for(int k=0; k<m; k++)
            {
                if(y.m[j][k]==0) continue;
                temp.m[i][k]+=x.m[i][j]*y.m[j][k];
            }
        }
    }
    return temp;
}
bool judge(const matrix &x,const matrix &y)
{
    for(int i=0;i<m;i++)
    for(int j=0;j<m;j++)
    if(x.m[i][j]!=y.m[i][j])
    return false;
    return true;
}
void buildMap()
{
    init();
    if(n==0||m==0)
    return ;
    for(int i=0;i<n;i++)
    for(int j=0;j<n;j++)
    if(i!=j)
    {
        matrix temp=multiply(M[i],M[j]);
        for(int e=0;e<n;e++)
        if(e!=i&&e!=j&&judge(temp,M[e]))
        map[i][e]=1;
    }
}
void spfa()
{
    int inq[N]={0},s;
    queue<int>q;
    for(int i=0;i<n;i++)
    {
        dis[i][i]=0;
        q.push(i);
        while(!q.empty())
        {
            s=q.front(); q.pop();
            inq[s]=0;
            for(int j=0;j<n;j++)
            if(map[s][j]&&dis[i][j]>dis[i][s]+1)
            {
                dis[i][j]=dis[i][s]+1;
                if(inq[j]==0)
                inq[j]=1,q.push(j);
            }
        }
    }
}
int main()
{
    int Q,a,b;
    while(scanf("%d%d",&n,&m)>0&&n+m!=0)
    {
        for(int i=0;i<n;i++)
        for(int j=0;j<m;j++)
        for(int e=0;e<m;e++)
        scanf("%d",&M[i].m[j][e]);

        buildMap();
        spfa();

        scanf("%d",&Q);
        while(Q--)
        {
            scanf("%d%d",&a,&b);
            a--; b--;
            if(dis[a][b]!=inf)
            printf("%d\n",dis[a][b]);
            else
            printf("Sorry\n");
        }
    }
    return 0;
}