HDU

题目：

The Shortest Path

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1827    Accepted Submission(s): 589

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

 

　　题意：给你n个m*m的矩阵，如果某个矩阵A*不是A的某个矩阵B=非A、B的矩阵C(这里的ABC只是编号上面的不一样)，那么就说明A到C有一条有向边，长度为1，然后给你q个询问，求询问中Ｘ到Ｙ的最短路，如果Ｘ到Ｙ没有路，就输出Sorry。

　　这一题的实质就是一个矩阵的乘法、矩阵的相等判定以及最短路，矩阵的乘法直接按照定义求，对于判定相等，可以直接逐个位置的元素判定是否相等也可以用memcmp()函数来判断，不过如果是用后者的话要注意空余的位置要在一开始赋值为0或者-1，总之就是排除干扰的因素，最短路这里用了Flyod。

　　这一题难度其实不是很大，就是做的时候有点累，结果敲完代码以后交上去一直WA，后来检查代码才发现Flyod写好了但是没有调用= =。

　　看来要注意合理休息。

 

代码：

 

#include <cstdio>
#include <cstring>
#define INF 1000000000
#define MAX 80
using namespace std;

int n,m,dis[MAX][MAX];

typedef struct Mat
{
    int city[MAX][MAX];

    void Init()
    {
        memset(city,0,sizeof(city));
    }

    void Insert()
    {
        int i,j;
        Init();
        for(i=0;i<m;i++)
        {
            for(j=0;j<m;j++)
            {
                scanf("%d",&city[i][j]);
            }
        }
    }

    void print()
    {
        int i,j;
        for(i=0;i<m;i++)
        {
            for(j=0;j<m;j++)
            {
                printf("%d ",city[i][j]);
            }
            printf("
");
        }
    }

    Mat operator * (const Mat &b)
    {
        int i,j,u;
        Mat c;
        c.Init();
        for(u=0;u<m;u++)
        {
            for(i=0;i<m;i++)
            {
                if(!city[i][u]) continue;
                for(j=0;j<m;j++)
                {
                    c.city[i][j]+=city[i][u]*b.city[u][j];
                }
            }
        }
        return c;
    }

    bool operator == (const Mat &b)
    {
        return (memcmp(city,b.city,sizeof(city))==0);
        /*
        int i,j;
        for(i=0;i<m;i++)
        {
            for(j=0;j<m;j++)
            {
                if(city[i][j]!=b.city[i][j]) return 0;
            }
        }
        return 1;
        */
    }
}Mat;

Mat mat[1000];

void Flyod()
{
    int i,j,u;
    for(u=1;u<=n;u++)
    {
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=n;j++)
            {
                int len=dis[i][u]+dis[u][j];
                if(dis[i][j]>len) dis[i][j]=len;
            }
        }
    }

}

int main()
{
    int i,j,u,q,a,b;
    //freopen("data.txt","r",stdin);
    while(scanf("%d %d",&n,&m),(n+m))
    {
        for(i=1;i<=n;i++)
        {
            mat[i].Insert();
            //mat[i].print();
        }
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=n;j++)
            {
                dis[i][j]= i==j ? 0 : INF;
            }
        }
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=n;j++)
            {
                if(i==j) continue;
                Mat c=mat[i]*mat[j];
                for(u=1;u<=n;u++)
                {
                    if(i==u || j==u) continue;
                    if(c==mat[u]) dis[i][u]=1;
                }
            }
        }
        Flyod();
        scanf("%d",&q);
        for(i=0;i<q;i++)
        {
            scanf("%d %d",&a,&b);
            if(dis[a][b]<INF) printf("%d
",dis[a][b]);
            else printf("Sorry
");
        }
    }
    return 0;
}