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 21 12 21 11 12 24 411 33 21 12 21 11 12 24 311 30 0

 

Sample Output

1Sorry

 

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;}