hdu2807

The Shortest Path

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1525    Accepted Submission(s): 475

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

 

Recommend

lcy

 

只会暴力~~~~1500ms水过~

#include <iostream>
#include <cstdio>
#include <cstring>
#define inf 99999999
using namespace std;
int n,m;
int e[100][100][100];
int map[100][100];
int temp[100][100];
void floyd()
{
    int j,i,k;
    for(k=1;k<=n;k++)
        for(i=1;i<=n;i++)
            for(j=1;j<=n;j++)
            {
                if(map[i][j]>map[i][k]+map[k][j])
                    map[i][j]=map[i][k]+map[k][j];
            }
}
void getmap()
{
    int i,j,k;
    memset(e,0,sizeof(e));
    for(i=1;i<=n;i++)
        for(j=1;j<=m;j++)
            for(k=1;k<=m;k++)
                scanf("%d",&e[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;
        }
    //printf("123\n");
    for(i=1;i<=n;i++)
    {
        for(j=1;j<=n;j++)
        {
            if(i==j)
                continue;
            memset(temp,0,sizeof(temp));
            int l1,l2,l3;
            for(l1=1;l1<=m;l1++)
                for(l2=1;l2<=m;l2++)
                {
                    temp[l1][l2]=0;
                    for(l3=1;l3<=m;l3++)
                        temp[l1][l2]+=e[i][l1][l3]*e[j][l3][l2];
                }
            for(l1=1;l1<=n;l1++)
            {
                if(l1==i||l1==j)
                    continue;
                int flag=1;
                for(l2=1;l2<=m;l2++)
                {
                    for(l3=1;l3<=m;l3++)
                    {
                        if(e[l1][l2][l3]!=temp[l2][l3])
                        {
                            flag=0;
                            break;
                        }
                    }
                    if(flag==0)
                        break;
                }
                if(flag)
                    map[i][l1]=1;
            }
        }
    }
    //printf("345\n");
    floyd();
}
int main()
{
    while(scanf("%d%d",&n,&m))
    {
        if(n==0&&m==0)
            break;
        getmap();
        int k;
        scanf("%d",&k);
        int i;
        for(i=0;i<k;i++)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            if(map[a][b]>=inf)
                printf("Sorry\n");
            else
                printf("%d\n",map[a][b]);
        }
    }
    return 0;
}