HDU 2807 The Shortest Path【矩阵的快速比较】

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

分析： 如果用普通的乘法 很容易超时 可以把二维矩阵转化为一维矩阵。

#include<stdio.h>
#include<string.h>
#define clr(x)memset(x,0,sizeof(x))
#define inf 0x1f1f1f1f
int dis[82][82];
int mat[82][82][82];
int mac[82][82];
int n,m;
int ok(int c)
{
    int i;
    for(i=1;i<=m;i++)
        if(mac[0][i]!=mac[c][i])
            return 0;
    return 1;        
}
void mul(int a,int b)
{
    int i,j;
    for(i=1;i<=m;i++)
    {
        mac[0][i]=0;
        for(j=1;j<=m;j++)
            mac[0][i]+=mat[a][i][j]*mac[b][j];
    }
    for(i=1;i<=n;i++)
    {
        if(i==a||i==b)continue;
        if(ok(i))
            dis[a][i]=1;
    }
}
int main()
{
    int i,j,k,p,q,len;
    while(scanf("%d%d",&n,&m)&&(n&&m))
    {
        memset(dis,inf,sizeof(dis));
        for(i=1;i<=n;i++)
            for(j=1;j<=m;j++)
            {
                mac[i][j]=0;
                for(k=1;k<=m;k++)
                {
                    scanf("%d",&mat[i][j][k]);
                    mac[i][j]+=mat[i][j][k]*k;
                }
            }
        for(i=1;i<=n;i++)
            for(j=1;j<=n;j++)
            {
                if(j==i)continue;
                mul(i,j);
            }
        for(k=1;k<=n;k++)
            for(i=1;i<=n;i++)
            {
                if(dis[i][k]!=inf)
                for(j=1;j<=n;j++)
                    if(dis[i][k]+dis[k][j]<dis[i][j])
                        dis[i][j]=dis[i][k]+dis[k][j];
            }
        scanf("%d",&k);
        while(k--)
        {
            scanf("%d%d",&p,&q);
            if(dis[p][q]!=inf)
                printf("%d\n",dis[p][q]);
            else printf("Sorry\n");
        }
    }
    return 0;
}