HDU2807：The Shortest Path(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
   
   
   
   

 

 

题意：城市一共有n个地点，每个地点是一个m*m的矩阵，当a*b=c时，说明a可以到达c，是单向边

思路：这道题一开始还真担心超时，虽然说只有50个点，但是在计算矩阵乘积的时候复杂度可不低，并且我也确实超时几次了，我很疑惑一开始将getmp函数写在主函数里面超时，但是作为调用函数就不超时了，这是为什么呢？

 

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
#define inf  99999999

int n,m;
int map[85][85];
int a[85][85][85],tem[85][85];

void Floyd()
{
    int i,j,k;
    for(i = 1; i<=n; i++)
        for(j = 1; j<=n; j++)
            for(k = 1; k<=n; k++)
                if(map[j][i]+map[i][k]<map[j][k])
                    map[j][k] = map[j][i]+map[i][k];
}

void getmap()
{
    int i,j,k,x,y,z;
    for(i = 1; i<=n; i++)
            for(j = 1; j<=m; j++)
                for(k = 1; k<=m; k++)
                    scanf("%d",&a[i][j][k]);
        for(i = 0; i<=n; i++)
        {
            for(j = 0; j<=n; j++)
            {
                if(i == j)
                    map[i][j] = 0;
                else
                    map[i][j] = inf;
            }
        }
        for(i = 1; i<=n; i++)
        {
            for(j = 1; j<=n; j++)
            {
                if(i==j)
                    continue;
                memset(tem,0,sizeof(tem));
                for(x = 1; x<=m; x++)//矩阵相乘
                {
                    for(y = 1; y<=m; y++)
                    {
                        tem[x][y] = 0;
                        for(z = 1; z<=m; z++)
                        {
                            tem[x][y]+=a[i][x][z]*a[j][z][y];
                        }
                    }
                }
                for(x = 1; x<=n; x++)
                {
                    if(x == i || x == j)
                        continue;
                    int flag = 1;
                    for(y = 1; y<=m; y++)
                    {
                        for(z=1; z<=m; z++)
                        {
                            if(tem[y][z]!=a[x][y][z])
                            {
                                flag = 0;
                                break;
                            }
                        }
                        if(!flag)
                            break;
                    }
                    if(flag)
                        map[i][x] = 1;
                }
            }
        }
        Floyd();
}

int main()
{
    int i,k,j,x,y,z;
    while(scanf("%d%d",&n,&m))
    {
        if(n==0 && m == 0)
            break;
        getmap();
        scanf("%d",&z);
        while(z--)
        {
            scanf("%d%d",&x,&y);
            if(map[x][y] == inf)
                printf("Sorry\n");
            else
                printf("%d\n",map[x][y]);
        }
    }

    return 0;
}