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 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

 

Recommend

lcy

 

只会暴力~~~~1500ms水过~

#include <iostream>#include <cstdio>#include <cstring>#define inf 99999999using 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;}